Page 160 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 10] STOICHIOMETRY 149

EXAMPLE 10.20. What rise in temperature will occur if 24.5 kJ of heat is added to 175 g of a dilute aqueous solution of
sodium chloride [c = 4.10 J/g· C)] (a) by heating with a bunsen burner and (b) by means of a chemical reaction?
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Ans. (a and b) The source of the heat does not matter; the temperature rise will be the same in either case. Watch out for
the units!
heat 24 500 J
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t = = = 34.1 C
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(m)(c) (175 g)[4.10 J/(g· C)]
EXAMPLE 10.21. Calculate the heat of reaction per mole of water formed if 0.0500 mol of HCl and 0.0500 mol of NaOH
are added to 15.0 g of water, all at 18.0 C. The solution formed is heated from 18.0 C to 54.3 C. The specific heat of the
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solution is 4.10 J/(g· C).
Ans. The law of conservation of mass allows us to calculate the mass of the solution:
15.0g + 1.82 g + 2.00 g = 18.8g
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Heat = mc t = (18.8g)[4.10 J/(g· C)](36.3 C) = 2800 J
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2.80 kJ/(0.0500 mol water formed) = 56.0 kJ/mol
Solved Problems

MOLE-TO-MOLE CALCULATIONS
10.1. Can the balanced chemical equation dictate to a chemist how much of each reactant to place in a reaction
vessel?
Ans. The chemist can put in as little as is weighable or as much as the vessel will hold. For example, the fact that
a reactant has a coefficient of 2 in the balanced chemical equation does not mean that the chemist must put
2 mol into the reaction vessel. The chemist might decide to add the reactants in the ratio of the balanced
chemical equation, but that is not required. And even in that case, the numbers of moles of each reactant
might be twice the respective coefficients or one-tenth those values, etc. The equation merely states the
reacting ratio.

10.2. How many factor labels can be used corresponding to each of the following balanced equations?

(a)2 K + Cl 2 −→ 2 KCl
(b) NCl 3 + 3H 2 O −→ 3 HOCl + NH 3
Ans. (a)6:
2 mol K 2 mol K 1 mol Cl 2 1 mol Cl 2 2 mol KCl 2 mol KCl
1 mol Cl 2 2 mol KCl 2 mol K 2 mol KCl 2 mol K 1 mol Cl 2
(b) 12: Each of the four compounds as numerators with the three others as denominators—4 × 3 = 12.

10.3. Which of the factors of Problem 10.2a would be used to convert (a) the number of moles of Cl 2 to the
number of moles of KCl, (b)KtoCl 2 , and (c)Cl 2 to K?
2 mol KCl 1 mol Cl 2 2 mol K
Ans. (a) (b) (c)
1 mol Cl 2 2 mol K 1 mol Cl 2
10.4. How many moles of AlCl 3 can be prepared from 7.5 mol Cl 2 and sufficient Al?
Ans. 2Al + 3Cl 2 −→ 2 AlCl 3

2 mol AlCl 3
7.5 mol Cl 2 = 5.0 mol AlCl 3
3 mol Cl 2
10.5. How many moles of H 2 O will react with 2.25 mol PCl 5 to form HCl and H 3 PO 4 ?
Ans. 4H 2 O + PCl 5 −→ 5 HCl + H 3 PO 4
4 mol H 2 O

2.25 mol PCl 5 = 9.00 mol H 2 O
1 mol PCl 5