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Chemistry 102 – Chapter 3 Summary and Review Answers

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Chemistry 102 – Chapter 3 Summary and Review Answers Chemistry 102 – Chapter 3 Summary and Review Answers 1. How are one mole of hydrogen gas and helium gas similar and how are they different? Answer: Within the context of Chapter 2 – one mole of hydrogen gas and one mole of helium gas contains the same number of gas particles. However, these particles for hydrogen are molecules (containing two atoms per molecule) and the particles for helium are atoms. Also, one mole of hydrogen molecules contains atoms with different composition and subsequently a different combined mass (2 g) compared to one mole of helium (4 g). 2. How are the states of matter denoted in a balanced chemical reaction? Answer: solid = s liquid = l gas = g aqueous = aq 3. Of formula units (or molecules), moles and mass – what can directly be compared for two different substances in a balanced chemical equation? Answer: Formula units and moles can be compared directly compared using stoichiometric coefficients from a balanced chemical equation. 4. Gallium has two stable isotopes, gallium‐69 and gallium‐71. If the mass of gallium‐69 is 68.925581 amu and is 60.108% abundant, what is the mass of gallium‐71? Answer: Because there are only two stable isotopes of gallium, the relative abundance of the two must combine to equal 100% or the relative abundance of gallium‐71 = 100% ‐ 60.81%. To organize the information: 69Ga 68.925581 amu 60.81% 71Ga to be determined 100% – 60.81% The average atomic mass is the weighted average of all of the stable isotopes: 69 71 69 69 ⎛⎞% Ga69 ⎛⎞ % Ga 71 ⎛⎞ % Ga 69 ⎛ 100%− % Ga ⎞ 71 Ga=+=+⎜⎟() Ga ⎜⎟() Ga ⎜⎟() Ga ⎜ ⎟() Ga ⎝⎠100% ⎝⎠ 100% ⎝⎠ 100% ⎝ 100% ⎠ Solving for the mass of gallium‐71: 69 ⎛⎞%Ga 69 ⎛⎞60.108% Ga− ⎜⎟() Ga 69.723 amu− ⎜⎟() 68.925581 amu 100% 100% 71Ga==⎝⎠ ⎝ ⎠ = 70.925 amu ⎛⎞100%− %69 Ga ⎛100%− 60.108% ⎞ ⎜⎟ ⎜⎟ ⎝⎠100% ⎝⎠100% 5. Show the balanced chemical equation for the reaction of ammonia with oxygen to form nitrogen monoxide and water using names, formula and particles. Answer: Names: Ammonia gas + oxygen gas →nitrogen monoxide gas + liquid water Formula: 4NH3(g) + 5O2(g) →4NO(g) + 6H2O(l) Particles: (Took liberties with representing the state of water – all of these molecules are very close together as a gas but that being said, water is represented here as a gas) 6. For the synthesis of ammonia from its elements, suppose the reaction yields 62.5%. What mass of hydrogen is needed in excess nitrogen to experimentally produce 1250 g of ammonia? Answer: First, you must balance the chemical equation: 3H2(g) + N2(g) → 2NH3(g) Now recognize that the experimental yield and percent yield will give the theoretical yield. This will then be used to determine the amount of the reactant needed. ⎛⎞100 g NH() theo yield ⎛⎞⎛⎞mol NH 3 mol H⎛⎞ 2.02 g H 1250 g NH exp yield3 3 22= 356 g H ()3 ()⎜⎟⎜⎟⎜⎟⎜⎟2 ⎝⎠62.5 g NH3332 ()exp yield⎝⎠⎝⎠ 17.04 g NH 2 mol NH⎝⎠ mol H 7. A 9.875 g sample of fuel is combusted in excess oxygen and 21.69 g of carbon dioxide and 11.84 g of water are collected. What is the empirical formula of the fuel? Answer: First determine the mass of carbon in the fuel and hydrogen in the fuel to determine if there is oxygen in the fuel: The mass of carbon in the fuel is found by the mass of carbon dioxide recovered: ⎛⎞⎛⎞mol CO2 1 mol C⎛⎞ 12.01 g C ()21.69 g CO2 ⎜⎟⎜⎟⎜⎟= 5.919 g C ⎝⎠⎝⎠44.01 g CO22 1 mol CO⎝⎠ mol C The mass of hydrogen in the fuel is found by the mass of water recovered: ⎛⎞⎛⎞mol H2 O 2 mol H⎛⎞ 1.01 g H ()11.84 g H2 O⎜⎟⎜⎟⎜⎟= 1.33 g H ⎝⎠⎝⎠18.02 g H22 O 1 mol H O⎝⎠ mol H The sum of the carbon and hydrogen is 7.25 g – less than 9.875 g, so there is oxygen in the fuel. Mass fuel = mass of carbon + mass of hydrogen + mass of oxygen Mass of oxygen = 9.875 g – 5.919 g – 1.33 g H = 2.63 g Now, determine the empirical formula by determining the number of moles of each element: ⎛⎞mol C⎛⎞ mol H ()5.919 g C⎜⎟== 0.493 mol C() 1.33 g H⎜⎟ 1.32 mol H ⎝⎠12.01 mol C⎝⎠ 1.01 g H ⎛⎞mol O ()2.63 g O⎜⎟= 0.164 mol O ⎝⎠16.00 g O CHO0.493 1.32 0.164 CHO 0.493 1.32 0.164 CHO 3 8 0.164 0.164 0.164 8. Show the chemical equation for the reaction of oxygen and hydrogen to make water describing which bonds break and which bonds form. Of the name, formula or particle representation for a chemical equation, which is best to describe which bonds break and which bonds form? Answer: Names: Hydrogen gas + oxygen gas →liquid water Formula: 2H2(g) + O2(g) → 2H2O(l) Particle representation: It is obvious that the bonds which must be broken are between the oxygen atoms in the oxygen molecules and between the hydrogen atoms in the hydrogen molecules. Similarly, the bonds that form are between hydrogen and oxygen and that two bonds form for every water molecule made. This is only obvious in the particulate representation (as would be with the structural formula as well). 9. When 50.0 g of oxygen reacts with 10.0 g of acetylene (C2H2) to produce carbon dioxide and water, what is the theoretical yield (in mass of carbon dioxide) and what is the limiting reactant? If the percent yield is 88.5%, what is the experimental yield? Answer: First, you must balance the chemical equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) You can calculate the theoretical yield and limiting reactant by calculating the mass of carbon dioxide produced from either reactant. The smaller mass will be the theoretical yield and the reactant producing this is the limiting reactant. ⎛⎞⎛⎞⎛⎞mol O222 4 mol CO 44.01 g CO ()50.0 g O2 ⎜⎟⎜⎟⎜⎟= 55.0 g CO2 ⎝⎠⎝⎠⎝⎠32.00 g O22 5 mol O mol CO 2 ⎛⎞⎛⎞⎛⎞mol C22 H 4 mol CO 2 44.01 g CO 2 ()10.0 g C22 H⎜⎟⎜⎟⎜⎟= 33.8 g CO2 ⎝⎠⎝⎠⎝⎠26.04 g C22 H 2 mol C 22 H mol CO 2 Now, using the percent yield, you can calculate the experimental yield. ⎛⎞88.5 g exp ()33.8 g theo⎜⎟= 29.9 g CO2 ⎝⎠100 g theo .
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