Page 162 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 10] STOICHIOMETRY 151

number of moles of phosphoric acid? (d) How many grams of sodium phosphate will be produced?
(e) How many moles of sodium hydroxide will it take to react with this quantity of phosphoric acid?
( f) How many grams of sodium hydroxide will be used up?
Ans. (a)H 3 PO 4 + 3 NaOH −→ Na 3 PO 4 + 3H 2 O

1 mol H 3 PO 4
(b)50.0gH PO 4 = 0.510 mol H 3 PO 4
3
98.0gH PO 4
3

1 mol Na 3 PO 4
(c) 0.510 mol H 3 PO 4 = 0.510 mol Na 3 PO 4
1 mol H 3 PO 4

164gNa PO 4
3
(d)0.510 mol Na 3 PO 4 = 83.6gNa PO 4
3
1 mol Na 3 PO 4
3 mol NaOH

(e) 0.510 mol H 3 PO 4 = 1.53 mol NaOH
1 mol Na 3 PO 4
40.0 g NaOH

( f )1.53 mol NaOH = 61.2 g NaOH
1 mol NaOH
10.14. (a) Write the balanced chemical equation for the reaction of sodium with chlorine. (b) How many moles
of Cl 2 are there in 7.650 g chlorine? (c) How many moles of NaCl will that number of moles of chlorine
produce? (d) What mass of NaCl is that number of moles of NaCl?
Ans. (a)2 Na + Cl 2 −→ 2 NaCl

1 mol Cl 2
(b)7.650gCl 2 = 0.1079 mol Cl 2
70.90gCl 2

2 mol NaCl
(c) 0.1079 mol Cl 2 = 0.2158 mol NaCl
1 mol Cl 2

58.45 g NaCl
(d)0.2158 mol NaCl = 12.61 g NaCl
1 mol NaCl
10.15. How many formula units of sodium hydroxide, along with H 2 SO 4 , does it take to make 7.50 × 10 22
formula units of Na 2 SO 4 ?
Ans. 2 NaOH + H 2 SO 4 −→ Na 2 SO 4 + 2H 2 O
23
22 1 mol Na 2 SO 4 2 mol NaOH 6.02 × 10 units NaOH
7.50 × 10 units Na 2 SO 4
23
6.02 × 10 units Na 2 SO 4 1 mol Na 2 SO 4 1 mol NaOH
23
= 1.50 × 10 units NaOH
Since the balanced chemical equation also relates the numbers of formula units of reactants and products,
the problem can be solved by converting directly with the factor label from the balanced equation:

2 units NaOH
22 23
7.50 × 10 units Na 2 SO 4 = 1.50 × 10 units NaOH
1 unit Na 2 SO 4
10.16. Howmanygramsofbariumhydroxidewillbeusedupinthereactionwithhydrogenchloride(hydrochloric
acid) to produce 16.70 g of barium chloride plus some water?
Ans. Ba(OH) 2 + 2 HCl −→ BaCl 2 + 2H 2 O

1 mol BaCl 2 1 mol Ba(OH) 2 171.3gBa(OH) 2
16.70 g BaCl 2 = 13.74gBa(OH) 2
208.2 g BaCl 2 1 mol BaCl 2 1 mol Ba(OH) 2
10.17. Draw a figure like Fig. 10-2 for Problem 10.15.
Ans. See Fig. 10-4.

10.18. (a) What reactant may be treated with phosphoric acid to produce 6.00 mol of potassium hydrogen
phosphate (plus some water)? (b) How many moles of phosphoric acid will it take? (c) How many moles
of the other reactant are required? (d) How many grams?