Page 164 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 10] STOICHIOMETRY 153

Ans. 3Ba(OH) 2 + 2H 3 PO 4 −→ Ba 3 (PO 4 ) 2 + 6H 2 O

1 mol H 3 PO 4 3 mol Ba(OH) 2 171gBa(OH) 2
14.7gH 3 PO 4 = 38.5gBa(OH) 2
98.0gH 3 PO 4 2 mol H 3 PO 4 1 mol Ba(OH) 2
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10.23. Calculate the number of moles of NaOH required to remove the SO 2 from 3.50 metric tons (3.50×10 g)
of atmosphere if the SO 2 is 0.10% by mass. (Na 2 SO 3 and water are the products.)
Ans. 2 NaOH + SO 2 −→ Na 2 SO 3 + H 2 O

0.10gSO 1 mol SO 2 2 mol NaOH
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3.50 × 10 g atm 2 = 110 mol NaOH
100 g atm 64.0gSO 2 1 mol SO 2
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10.24. Calculate the number of moles of Al 2 O 3 needed to prepare 4.00 × 10 g of Al metal in the Hall process:
electricity
Al 2 O 3 + 3C −−−−−−→ 2Al + 3CO
special solvent

1 mol Al 1 mol Al 2 O 3 4

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Ans. 4.00 × 10 gAl = 7.41 × 10 mol Al 2 O 3
27.0gAl 2 mol Al
10.25. How much AgCl can be prepared with 50.0 g CaCl 2 and excess AgNO 3 ?
Ans. CaCl 2 + 2 AgNO −→ Ca(NO 3 ) 2 + 2 AgCl
3

1 mol CaCl 2 2 mol AgCl 143 g AgCl
50.0 g CaCl 2 = 129 g AgCl
111 g CaCl 2 1 mol CaCl 2 1 mol AgCl
10.26. In chemistry recitation, a student hears (incorrectly) the instructor say, “Hydrogen chloride reacts with
Ba(OH) 2 ” when the instructor has actually said, “Hydrogen fluoride reacts with Ba(OH) 2 .” The instructor
then asked how much product is formed. The student answers the question correctly. Which section, 10.1
or 10.2, were they discussing? Explain.

Ans. They were discussing Sec. 10.1. Since the student got the answer correct despite hearing the wrong name,
they must have been discussing the number of moles of reactants and products. The numbers of moles of
HF and HCl would be the same in the reaction, but since they have different formula masses, their masses
would be different.
10.27. Consider the equation

KMnO 4 + 5 FeCl 2 + 8 HCl −→ MnCl 2 + 5 FeCl 3 + 4H 2 O + KCl

How many grams of FeCl 3 will be produced by the reaction of 2.72 g of KMnO 4 ?
Ans. The reacting ratio is given by the coefficients. The coefficients of interest are 1 for KMnO 4 and 5 for FeCl 3 .

1 mol KMnO 4 5 mol FeCl 3 162 g FeCl 3
2.72 g KMnO 4 = 13.9 g FeCl 3
158 g KMnO 4 1 mol KMnO 4 1 mol FeCl 3
10.28. How many grams of NaCl can be produced from 7.650 g chlorine?
Ans. This problem is the same as Problem 10.14. Problem 10.14 was stated in steps, and this problem is not, but
you must do the same steps whether or not they are explicitly stated.
10.29. How much KClO 3 must be decomposed thermally to produce 14.6 g O 2 ?
Ans. 2 KClO 3 −→ 2 KCl + 3O 2

1 mol O 2 2 mol KClO 3 122.6 g KClO 3
14.6gO 2 = 37.3 g KClO decomposed
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32.0gO 2 3 mol O 2 1 mol KClO 3