For Each Reaction Balance the Equation First

Stoichiometry and Balancing Equations

For each reaction balance the equation first 1) Determine the mass of lithium hydroxide, LiOH, produced when 0.38 g of lithium nitride, Li3N, reacts with water according to the following equation:

Li3N + 3H2O  NH3 + 3LiOH

0.38 g x 1 moles Li3N = 0.0109 moles Li3N 34.833 g

0.0109 moles Li3N x 3 moles LiOH = 0.0327 mole LiOH 1 mole Li3N

0.0327 mole LiOH x 23.941 g = 0.79 g LiOH 1 mole LiOH

2) What mass of sodium chloride, NaCl, is produced when chlorine reacts with 0.29 g of sodium iodide, NaI?

Cl2 + 2NaI  2NaCl + I2

0.29 g x 1 moles NaI = 0.001934 moles NaI 149.9 g

0.001934 moles NaI x 2 moles NaCl = 0.001934 mole NaCl 2 mole NaI

0.001934 mole NaCl x 58.45 g = 0.11 g NaCl 1 mole NaCl

3) Determine the mass of carbon dioxide, CO2, produced when 0.85 g of butane, C4H10, reacts with oxygen according to the following equation: (Hint to balance you will have to multiply by 2) 2C4H10 + 13O2  8CO2 + 10H20

0.85 g x 1 moles C4H10 = 0.01465 moles C4H10 58 g

0.01465 moles C4H10 x 8 mole CO2 = 0.05862 mole CO2 2 mole C4H10

0.05862 mole C4H10 x 44 g = 2.6 g CO2 1 mole CO2

4) Determine the mass of antimony, Sb, produced when 0.46 g of antimony(III) oxide, Sb2O3, reacts with carbon according to the following equation:

Sb2O3 + 3C 2Sb + 3CO

0.46 g x 1 moles Sb2O3.01465 = 0.0015775 mole Sb2O3 291.6 g

0.0015775 mole Sb2O3 x 2 mole Sb = 0.003155 mole Sb 1 mole Sb2O3

0.003155 mole Sb x 121.8 g = 0.38 g Sb 1 mole Sb

5) What mass of hydrogen peroxide (H2O2) must decompose to produce 0.77 g of water?

2H2O2  2H2O + O2

0.77 g x 1 moles H2O = 0.042777 mole H2O 18.02 g

0.042777 mole H2O x 2 mole H2O2 = 0.042777 mole H2O2 2 mole H2O

0.042777 mole H2O2 x 34.04 g = 1.5 g H2O2 1 mole H2O2 6) What mass of carbon monoxide, CO, must react with oxygen to produce 0.69 g of carbon dioxide, CO2?

2CO + O2  2CO2

0.69 g x 1 moles CO2 = 0.01568 mole CO2 44 g

0.01568 mole CO2 x 2 mole CO = 0.01568 mole CO 2 mole CO2

0.01568 mole CO x 28 g = 0.44 g CO 1 mole CO

7) Determine the mass of sodium nitrate, NaNO3, produced when 0.73 g of nickel (II) nitrate, Ni(NO3)2 , reacts with sodium hydroxide according to the following equation:

Ni(NO3)2 + 2NaOH Ni(OH)2 +2NaNO3

0.73 g x 1 moles Ni(NO3)2 = 0.003998 mole Ni(NO3)2 182.69 g

0.003998 Ni(NO3)2 mole x 2 mole NaNO3 = 0.00799 mole NaNO3 1 mole Ni(NO3)2

0.00799 mole NaNO3 x 85 g = 0.68 g NaNO3 1 mole NaNO3

8) Determine the mass of calcium hydroxide, Ca(OH)2, produced when calcium carbide reacts with 0.64 g of water, H2O, according to the following equation:

CaC2 + 2H2O  Ca(OH)2 + C2H2 0.64 g x 1 moles H2O = 0.0355 mole H2O 18 g

0.0355 H2O mole x 1 mole Ca(OH)2 = 0.01778 mole Ca(OH)2 2 mole H2O

0.01778 mole Ca(OH)2 x 74.1 g = 1.3 g Ca(OH)2 1 mole Ca(OH)2

9) How many grams of ozone (O3) must decompose to produce 0.87 g of oxygen?

2O3  3O2

0.87 g x 1 moles O2 = 0.0272 mole H2O 32 g

0.0272 O2 mole x 2 mole O3 = 0.018125 mole O3 3 mole O2

0.018125 mole O3 x 48 g = 0.87 g O3 1 mole O3

10) Find the mass of sugar (C6H12O6) required to produce 1.82 L of carbon dioxide, CO2, gas at STP from the reaction described by the following equation: (Remember 1 mole of gas = 22.4L at STP)

C6H12O6  2C2H6O +2CO2

1.82 L x 1 moles CO2 = 0.08125 mole CO2 22.4 L

0.08125 mole CO2 x 1 mole C6H12O6 = 0.040625 mole C6H12O6 2 mole CO2

0.040625 mole C6H12O6 x 180.18 g = 7.32 g C6H12O6 mole C6H12O6 11) How many liters of oxygen are necessary for the combustion of 425 g of sulfur, assuming that the reaction occurs at STP?

The balanced equation is S+ O2  SO2.

425 g x 1 moles S = 13.25226 mole CO2 32.07 g

13.25226 mole S x 1 mole O2 = 13.25226 mole O2 1 mole S

13.25226 mole O2 x 22.4 L = 297 L O2 1 mole O2

12) Find the mass of benzene (C6H6) required to produce 2.66 L of carbon dioxide, CO2, gas at STP from the reaction described by the following equation: (Hint to balance you will have to multiply by 2)

2C6H6 + 15O2  6H2O + 12CO2

2.66 L x 1 moles CO2 = 0.11875 mole CO2 22.4 L

0.11875 mole CO2 x 2 mole C6H6 = 0.01979 mole C6H6 12 mole CO2

0.01979 mole C6H6 x 78.12 g = 1.55 g C6H6 mole C6H6