Chemical Reactions and Quantities

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Chemical Reactions and Quantities Chapter 7 Chemical Reactions occur Everywhere… • …when fuel burns with oxygen in our cars to make the car move… • …when we cook our food… • …when we dye our hair… • …in our bodies, chemical reactions convert food into molecules that build and move muscle… • …in leaves of trees and plants, carbon dioxide and water are converted to carbohydrates… Some chemical reactions are simple, whereas others are quite complex… • However, they can all be written by chemical equations that chemists use to describe chemical reactions.

• In every chemical reaction, atoms are rearranged to give new substances. – Just like following a recipe, certain ingredients are combined (and often heated) to form something new. Chapter Seven

Write a balanced equation from formulas of the reactants and products for a reaction; determine the number of atoms in the reactants and products. Chemical Change

• A chemical change occurs when a substance is converted into one or more new substances that have different formulas and properties.

• For example, when silver tarnishes, the shiny, silver metal (Ag) reacts with sulfur (S) to become the dull, black substance we call tarnish

(Ag2S). Chemical Reaction

• A chemical reaction always involves a chemical change because atoms of the reacting substances form new combinations with new properties.

• For example, a chemical reaction (and chemical change) takes place when a piece of iron (Fe) combines with oxygen (O2) in the air to produce a new substance, rust (Fe2O3), which has a reddish- brown color. Evidence of a Chemical Reaction

During a chemical change, new properties are often visible, which indicates that a chemical reaction took place. Chemical Equation

When you build a model airplane or prepare a new recipe, you follow a set of direction.

These directions tell you what materials to use and the products you will obtain.

In chemistry, a chemical reaction tells us the materials we need and the products that will form. Writing a chemical equation

Suppose you work in a bicycle shop assembling wheels and frames into bicycles.

You could represent this by a simple equation: Writing a chemical equation

Return to the silver example:

When silver tarnishes, the shiny, silver metal (Ag) reacts with sulfur (S) to become the dull, black substance we call tarnish (Ag2S).

* Unbalanced equation Writing a chemical equation

Return to the iron example: A chemical reaction takes place when a piece of iron (Fe) combines with oxygen (O2) in the air to produce a new substance, rust (Fe2O3), which has a reddish-brown color.

Generally, each formula is followed by an abbreviation, in parentheses, that gives the physical state of the substance. solid (s) liquid (l) gas (g) aqueous (aq) dissolved in water * Unbalanced equation Writing a chemical equation

Some reactions require heat to be added in order for the change to occur.

For example, when you burn charcoal in a grill, the carbon (C) in the charcoal combines with the oxygen (O2) to form carbon dioxide.

* Unbalanced equation Chemical equation symbols

Cu(s) + S(s) → CuS(s)

Δ CaCO3(s) → CaO(s) + CO2(g)

Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq) Identifying a balanced chemical equation

When a chemical reaction takes place, the bonds between the atoms of the reactants are broken and new bonds are formed to give the products. Identifying a balanced chemical equation

All atoms are conserved which means that atoms cannot be gained, lost, or changed into another type of element during the chemical reactions

Every chemical reaction must be written as a balanced reaction, which shows the same number of atoms for each element for the reactants and in the products. Balanced chemical equation

Consider the following in which hydrogen (H2) reacts with oxygen (O2) to form water:

H2(g) + O2(g)  H2O(g) unbalanced there are the same number of hydrogens on both sides (2) but different numbers of oxygen atoms (2 and 1) so it is unbalanced. Balanced chemical equation

H2(g) + O2(g)  H2O(g) unbalanced

We use whole numbers called coefficients in front of formulas. Coefficients indicate how many of each molecule are in an equation.

In the balanced equation, there are two H2 molecules and two H2O molecules:

This illustrates the Law of Conservation of Matter which states that matter cannot be created or destroyed during a chemical reaction. Practice counting atoms

Indicate the number of each type of atom in the following balanced chemical equation:

Fe2S3(s) + 6HCl(aq)  2FeCl3(aq) + 3H2S(g)

reactants products Fe S H Cl Practice counting atoms

Indicate the number of each type of atom in the following balanced chemical equation:

Δ 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

reactants products C H O Balancing chemical equations

When silver tarnishes, the shiny, silver metal (Ag) reacts with sulfur (S) to become the dull, black substance we call tarnish

(Ag2S). What is the balanced chemical equations describing this reaction? Balancing chemical equations

A chemical reaction takes place when a piece of iron (Fe) combines with oxygen (O2) in the air to produce a new substance, rust (Fe2O3), which has a reddish-brown color. What is the balanced chemical equation describing this reaction? Balancing chemical equations

Balance the following chemical reaction:

Na3PO4(aq) + MgCl2(aq)  Mg3(PO4)2(s) + NaCl(aq) Chapter Seven

Identify a reaction as a combination, decomposition, single replacement, double replacement, or combustion. A great number of reactions occur in nature, in biological systems, and in the laboratory.

However, there are some general patterns among all reactions that help us classify reactions. * note – some reactions may fit in more than one category. Combination Reactions

• In a combination reaction, two or more elements or compounds bond to form one product.

• For example, sulfur and oxygen combine to form the product sulfur dioxide. Combination Reactions

2Mg(s) + O2(g)  2MgO(s)

N2(g) + 3H2(g)  2NH3(g)

Cu(s) + S(s)  CuS(s)

MgO(s) + CO2(g)  MgCO3(s) Decomposition Reactions

In a decomposition reaction, a reactant splits into two or more simpler products.

For example, when mercury (II) oxide is heated, the compound breaks apart into mercury atoms and oxygen. Δ 2HgO(s) → 2Hg(l) + O2(g) Decomposition Reactions

Δ CaCO3(s) → CaO(s) + CO2(g)

Fe2S3(s)  2Fe(s) + 3S(s) Replacement Reactions

In a replacement reaction, elements in a compound are replaced by other elements.

In a single replacement reaction, one element switches places with another element in the reactants.

*A and B switch places. Replacement Reactions

Zn(s) + 2HCl(aq)  H2(g) + ZnCl2(aq)

Cl2(g) + 2KBr(s)  2KCl(s) + Br2(l) Replacement Reactions

In a double replacement reaction, the positive ions in the reacting compounds switch places. Replacement Reactions

BaCl2(aq) + Na2SO4(aq) 

NaOH(aq) + HCl(aq)  Combustion Reactions

The burning of a candle and the burning of fuel in the engine of a car are examples of combustion reactions.

In a combustion reaction, a carbon-containing compound burns in oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) and energy in the form of heat or flame.

Fuel + O2  CO2 + H2O *unbalanced

Fuel often: methane (CH4), propane (C3H8), or similar. Sometimes also has oxygen atoms in the formula: C3H7O Combustion Reactions

Fuel + O2  CO2 + H2O unbalanced

Methane gas (CH4) aka natural gas undergoes combustion when used to cook food on a gas stove:

Δ CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + energy

Combustion of propane (C3H8):

Δ C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) + energy Summary of Reaction Types Practice

Classify as combination, decomposition, single replacement, double replacement, or combustion reaction.

2Fe2O3(s) + 3C(s)  3CO2(g) + 4Fe(s) Practice

Classify as combination, decomposition, single replacement, double replacement, or combustion reaction.

Δ 2KClO3(s) → 2KCl(s) + 3O2(g) Practice

Classify as combination, decomposition, single replacement, double replacement, or combustion reaction.

Δ C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g) + energy Chapter Seven

Define the terms oxidation and reduction, identify the reactants as oxidized and reduced. Another type of reaction is the:

oxidation-reduction reaction

This is a continuation of the previous section Types of Reactions. The oxidation-reduction reaction is arguably one of the most important and undeniably the most complicated of the types we will discuss in this chapter.

Because of this, oxidation-reduction reactions has it’s own section in this chapter. Perhaps you have never heard of an oxidation and reduction reaction. However, this type of reaction has many important applications in your everyday life.

▪ Rusty nail

▪ Tarnish on a silver spoon

▪ Corrosion on metal

▪ When you turn on the lights in your car, an oxidation-reduction reaction within the car battery provides the electricity.

▪ When you light a campfire, as the wood burns, oxygen combines with carbon and hydrogen to produce carbon dioxide, water, and heat. - In section 7.2 we called this a combustion reaction. It is. It’s also a oxidation-reduction reaction.

▪ When you eat starchy food, the starches break down to give glucose which is oxidized in our cells to give you energy. Oxidation-Reduction Reactions

In an oxidation-reduction reaction (redox), electrons are transferred from one substance to another. Oxidation-Reduction Reactions

If one substance loses an electron, another substance must gain the electron.

The substance that lost electrons was oxidized and we define oxidation as the loss of electrons

The substance that gained electrons was reduced and we define reduction as the gain of electrons Oxidation-Reduction Reactions

The substance that lost electrons was oxidized and we define oxidation as the loss of electrons

The substance that gained electrons was reduced and we define reduction as the gain of electrons Redox – Example 1

The green color that appears on copper surfaces is due partly to:

2Cu(s) + O2(g)  2CuO(s)

This is a redox reaction, and we can break the equation up to show this. Redox – Example 1 OIL RIG

Break the equation into oxidation and reduction.

2Cu(s) + O2(g)  2CuO(s)

Oxidation Going from charge 0 to +2, Cu lost two electrons and Cu was oxidized. Redox – Example 1 OIL RIG

Break the equation into oxidation and reduction.

2Cu(s) + O2(g)  2CuO(s)

Reduction

Going from charge 0 to -2, O2 gained four electrons and O2 was reduced. (Each O went from 0 to -2 and gained 2 electrons. Because there are

two oxygen atoms in O2, that’s a total of four electrons.) Redox – Example 1

To summarize:

2+ - 2Cu(s)  2Cu (s) + 4e oxidation

- 2- O2(g) + 4e  2O (s) reduction

In a redox reaction, there’s always two pieces that occur. One substance is oxidized (loses electrons) and another substance is reduced (gains the lost electrons). Redox – Example 1

Original equation:

2Cu(s) + O2(g)  2CuO(s)

The two half reactions can be added back together to give the original equation.

2+ - 2Cu(s)  2Cu (s) + 4e oxidation

- 2- O2(g) + 4e  2O (s) reduction Redox – Example 2 OIL RIG

Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)

Oxidation half-reaction:

Reduction half-reaction:

______was oxidized

______was reduced Redox – Example 3 OIL RIG

Zn(s) + Cl2(g)  ZnCl2(s)

Oxidation half-reaction:

Reduction half-reaction:

______was oxidized

______was reduced Chapter Seven

Use Avogadro’s number to determine the number of particles in a given number of moles. At the grocery store, you buy eggs by the dozen, or soda by the case. At Staples, pencils are ordered by the gross and paper by the ream.

The terms dozen, case, gross, and ream are used to count the number of items present.

Chemists do the same thing, and they call it the mole. The mole

In chemistry, particles (such as atoms, molecules, and ions) are counted by the mole which contains 6.02 x 1023 items. Avogadro’s number

6.02 x 1023 is known as Avogadro’s number.

602 000 000 000 000 000 000 000 = 6.02 x 1023

It is a very large number and we use it because atoms are so small that it takes an extremely large number of atoms to provide a usable amount to use in chemical reactions in a laboratory. The mole and Avogadro’s number

1 mole of any element always contains Avogadro’s number of atoms.

1 mole of carbon atoms contains 6.02 x 1023 carbon atoms. 1 mole of aluminum atoms contains 6.02 x 1023 aluminum atoms.

1 mole of any element = 6.02 x 1023 atoms of that element. 1 mole of any compound = 6.02 x 1023 of that compound

23 1 mole of CO2 contains 6.02 x 10 molecules of CO2. 1 mole of NaCl contains 6.02 x 1023 molecules of NaCl.

Converting mole ↔ particles

We use Avogadro’s number as a conversion factor to convert between the moles of a substance and the number of particles it contains.

6.02 x 1023 particles 1 mole 1 mole 6.02 x 1023 particles

The particles can be atoms, molecules, ions, or anything. Replace particles with items if that helps. Practice

Convert 4.00 moles of sulfur to atoms of sulfur. Practice

24 Convert 3.01 x 10 molecules of CO2 to moles of CO2 moles in chemical formulas

We have seen in past chapters that the subscripts in the chemical formula of a compound indicate the number of atoms of each type of element in the compound.

For example, aspirin, C9H8O4 moles in chemical formulas

The subscripts also tell us the number of moles of each element in 1 mole of aspirin. moles in chemical formulas

And because 1 mole aspirin = 6.02 x 1023 molecules of aspirin moles in chemical formulas

We can also use subscripts from the formula, C9H8O4, we can write the conversion factors for each of the elements in 1 mole of aspirin. Practice

Propyl acetate, C5H10O2, gives the odor and taste of pears. How many moles of C are present in 1.50 moles of propyl acetate? Chapter Seven

Calculate the molar mass for a substance given its chemical formula; use molar mass to convert between grams and moles. A single atom or molecule is much too small to weigh, even using the best scale.

In fact, it takes a huge number of atoms or molecules to make enough of a substance for you to see.

An amount of water that contains Avogadro’s number of water molecules (6.02 x 1023) is only a few sips!

So in a lab, we use a scale to weigh out substances in moles. Molar Mass

For any element, the quantity called molar mass is the quantity in grams that equals the atomic mass of that element. Molar Mass

We are counting 6.02 x 1023 atoms of an element when we weigh out the number of grams equal to the molar mass.

For example, carbon has an atomic mass of 12.01 on the periodic table. This means 1 mole of carbon atoms has the mass of 12.01g.

Then to obtain 1 mole of carbon atoms, we would weigh out 12.01g of carbon.

Atomic mass of Carbon: 12.01 amu Molar mass of Carbon: 12.01 g/mol Using Molar Mass

The molar mass of an element is useful to convert moles of an element to grams (or grams to moles).

For example, 1 mole of sulfur as a mass of 32.066g

1 mole S = 32.066g of S Molar Mass of a Compound

We can calculate the molar mass of a compound by adding up the molar mass of each atom.

Example: Calculate the molar mass of lithium carbonate (Li2CO3) Practice

Calculate the molar mass for salicyclic acid, C7H6O3. Examples of 1 mole Calculations using Molar Mass

We can now change from moles to grams (or grams to moles) using the molar mass as a conversion factor.

Example: Convert 0.750 moles of silver to grams of silver. Calculations using Molar Mass

We can change moles to grams (or vice versa) for a compound using the compound’s molar mass.

Example: A box of salt contains 737 g of NaCl. How many moles of NaCl are present? Summary Chapter Seven

7.1 – Equations for Chemical Reactions 7.2 – Types of Reactions 7.3 – Oxidation-Reduction Reactions 7.4 – The Mole 7.5 – Molar Mass and Calculations 7.6 – Mole Relationships in Chemical Equations 7.7 – Mass Calculations for Reactions 7.8 – Limiting Reactants and Percent Yield 7.9 – Energy in Chemical Reactions 7.6 Mole Relationships in Chemical Equations Give a quantity in moles of reactant or product, use a mole-mole factor from the balanced chemical equation to calculate the number of moles of another substance in the reaction. Law of Conservation of Mass

In any chemical reaction, the total amount of matter in the reactants is equal to the total amount of matter in the products. - an application of the Law of Conservation of Mass

Thus no material is lost or gained as original substances are changed into new substances. For example, tarnish (Ag2S), forms when silver reacts with sulfur to form silver sulfide:

2Ag(s) + S(s)  Ag2S(s)

For every 2 silver atoms, 1 sulfur atom is required Or For every 2 moles of silver atoms, 1 mole of sulfur atoms is required.

Because the molar mass can be determined, the moles of Ag, S, and

Ag2S can also be stated in terms of mass (grams) of each… Thus 215.8 g of Ag and 32.1 g of S reacts to form 247.9 g of Ag2S.

215.8g + 32.1g = 249.9g The total mass of reactants (249.9 g) equals the total mass of the products (249.9 g). Info in a balanced equation

The various ways in which a chemical equation can be interpreted: Mole-mole factors

When iron reacts with sulfur, the product is iron(III) sulfide:

2Fe(s) + 3S(s)  Fe2S3(s)

From the balanced equation, we see that 2 moles of iron reacts with 3 moles of sulfur to form mole of iron(II) sulfide.

(Actually, any amount of iron or sulfur may be used, but the ratio of iron reacting with sulfur will always be the same.) Mole-mole factors

From the coefficients, we can write mole-mole factors between any two compounds in an equation.

2Fe(s) + 3S(s)  Fe2S3(s)

Fe and S:

Fe and Fe2S:

S and Fe2S: Mole-mole factors in calculations

Whenever you prepare a recipe, you need to know the proper amounts of ingredients and how much the recipe will make.

The same is true for chemistry.

Now that we have written all the possible mole-mole factors for

2Fe(s) + 3S(s)  Fe2S3(s) we will use those mole-mole factors in a chemical calculation. Practice #1

How many moles of sulfur are needed to react with 1.42 moles of iron?

2Fe(s) + 3S(s)  Fe2S3(s)

2 moles Fe 2 moles Fe 3 moles S

3 moles S 1 mole Fe2S3 1 moleFe2S3 Practice #2

How many moles of iron are needed to react with 2.75 moles of sulfur?

2Fe(s) + 3S(s)  Fe2S3(s)

2 moles Fe 2 moles Fe 3 moles S

3 moles S 1 mole Fe2S3 1 moleFe2S3 Chapter Seven

Given the mass in grams of a substance in a reaction, calculate the mass in grams of another substance in the reaction. When we have the balanced chemical equation for a reaction, we can use the mass of one of the substances (A) to calculate the mass of a second (B).

However, the calculations require us to convert the mass of A to the moles of A (using A’s molar mass). Then we use the mole- mole factor to convert moles A to moles B. Then finally use B’s molar mass to convert moles B to grams B.

molar mole-mole molar mass A factor mass B grams A moles A mole B grams B Practice

How many grams of CO2 are produced when 54.6g of C2H2 is burned? Δ 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g) Practice

Calculate grams CO2 produced when 25 g of O2 reacts. Δ 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g) Chapter Seven

7.1 – Equations for Chemical Reactions 7.2 – Types of Reactions 7.3 – Oxidation-Reduction Reactions 7.4 – The Mole 7.5 – Molar Mass and Calculations 7.6 – Mole Relationships in Chemical Equations 7.7 – Mass Calculations for Reactions 7.8 – Limiting Reactants and Percent Yield 7.9 – Energy in Chemical Reactions 7.8 Limiting Reactants and Percent Yield Identify the limiting reactant and calculate the amount of product formed from the limiting reactant. Given the actual quantity of product, determine the percent yield for a reaction. Limiting Reactant

When you make peanut butter sandwiches for lunch, you need 2 slices of bread and 1 tablespoon of peanut butter for each sandwich. 1 Tbl. 2 slices peanut 1 peanut butter bread + butter sandwich If you have 8 slices of bread and a full jar of peanut butter, you will run out of bread after you make 4 peanut butter sandwiches once the bread is used up, even though there is a lot of peanut butter left in the jar. The number of slices of bread has limited the number of sandwiches you can make. Limiting Reactant

On a different day, you might have 8 slices of bread but only a tablespoon of peanut butter left in the peanut butter jar. You will run out of peanut butter after you make just 1 sandwich and have 6 slices of bread left over. The smaller amount of peanut butter available has the limited the number of sandwiches you can make. Limiting Reactant

The reactant that is completely used up is the limiting reactant.

The reactant that does not completely react and is left over is called the excess reactant. Moles from Limiting Reactant

In the same way, the reactants in a chemical reactions do not always combine in quantities that allow each to be used up at exactly the same time.

In many reactions, there is a limiting reactant that determines the amount of product that can be formed. Identifying Limiting Reactant

When we know the quantities of the reactants of a chemical reaction we calculate the amount of product that is possible from each reactant if it were completely consumed. The limiting reactant is the one that runs out first and produces the smaller amount of product. Practice #1

If 3.00 moles of CO and 5.00 moles of H2 are the initial reactants, what is the limiting reactant? And how many moles of CH3OH can be produced.

CO(g) + 2H2(g)  CH3OH(g) Practice #2

If 4.00 moles of CO and 4.00 moles of H2 are the initial reactants, what is the limiting reactant? And how many moles of CH3OH can be produced.

CO(g) + 2H2(g)  CH3OH(g) Mass from Limiting Reactant

The quantities of the reactants can also be given in grams. We combine knowledge from sections 7 and 8 to find the limiting reactant.

The calculation to find the limiting reactant is the same as the last two Practice slides. But before we can do it, we must convert grams to moles. - Subscripts in chemical equations are not grams!! - You need to use moles to compare subscripts!! Practice

How many grams of CO are formed from 70.0 g of SiO2 and 50.0 g of C? Δ SiO2(s) + 3C(s) → SiC(s) + 2CO(g) Percent Yield

In our reactions so far we have assumed 100% efficiency, meaning all of the reactants changed completely to products.

While this is ideal, it’s hardly reality. Often, reactant is lost when moving from container to container, chemicals aren’t 100% pure, and unwanted side reactions use up reactant making unwanted products. Thus 100% of the desired product is not actually obtained.

When we do a chemical reaction in the lab, we measure out specific quantities of the reactants. We then calculate the theoretical yield for the reaction. Which is the amount of product (100%) we would expect if all the reactants were converted to the desired products.

When the reaction ends, we collect and measure the mass of the product that was actually made. This is called the actual yield.

Because the reaction is hardly ever 100% efficient the actual yield is less than the theoretical yield. Percent Yield

We can represent the ratio of actual yield to theoretical yield in one number called percent yield.

actual yield % yield = x 100 theoretical yield Practice

What is the % yield of LiHCO3 for the reaction if 50.0 g of LiOH gives 72.8 g LiHCO3.

LiOH(s) + CO2(g)  LiHCO3(s) Chapter Seven

Given the heat of reaction, calculate the loss or gain of heat for an exothermic or endothermic reaction. Energy in Chemical Reactions

Almost every chemical reaction involves a loss or gain of energy.

To discuss energy change for a reaction, we look at the energy of the reactants before the reaction and the energy of the products after the reaction.

The SI unit for energy is the joule (J) 1 kilojoule (kJ) = 1000 joules (J) Heat of Reaction

Energy is often present as heat.

The heat of reaction is the amount of heat absorbed or released during a reaction that takes place at constant pressure.

A change of energy occurs as reactants interact, bonds break apart, and products form.

Another name for heat of reaction is enthalpy change.

ΔH = Hproducts - Hreactants Exothermic Reactions

In an exothermic reaction the energy of the products is lower than that of the reactants. - Heat is released along with the products that form.

For example when 1 mole of H2 and 1 mole of Cl2 react, 2 mole HCl form. The energy of the HCl is 185kJ less than the H2 and Cl2

Because products are lower energy, that energy difference is released as heat.

We write ΔH with a negative sign to indicate exothermic. Exo example: Hot Packs

Hot packs are used in medicine, camping gear, emergency kits, etc. Inside the hot pack are two pouches. One pouch has water, the other has CaCl2 When the boundary is broken between pouches an exothermic reaction takes place and releases heat.

퐻2푂 CaCl2(s) CaCl2(aq)

Temperatures can increase as much as 66C Endothermic Reactions

In an endothermic reaction, the energy of the products is higher than that of the reactants. - Heat is required to begin the reaction and convert reactants to products.

For example: 1 mole N2 and 1 mole O2 react to form 2 moles NO only when at least 180 kJ of heat is added.

We write ΔH with a positive sign to indicate endothermic. Endo example: Cold Packs

Cold packs work the same way with two pouches. One pouch as water, the other has NH4NO3. The temperature can drop to 4-5C.

퐻2푂 NH4NO3(s) + 26kJ NH4NO3(aq) Calculations of Heat in Reactions

The value of ΔH refers to the heat change for each substance in the balanced equation.

2H2O(l)  2H2(g) + O2(g) Δ H = +572 kJ

means that for this reaction, +572 kJ are absorbed by 2 moles of H2O to produce 2 moles of H2 an 1 mole of O2.

We can use ΔH to write conversion factors just like the mole-mole factors from section 7.6. Calculations of Heat in Reactions

Suppose that 9.00 g H2O undergoes the reaction. Calculate the heat absorbed.

2H2O(l)  2H2(g) + O2(g) Δ H = +572 kJ Practice

How much heat, in kJ, is released when nitrogen and hydrogen react to form

50.0g of ammonia (NH3)?

N2(g) + 3H2(g)  2NH3(g) Δ H = -92.2 kJ Chapter Seven