Page 165 - Theory and Problems of BEGINNING CHEMISTRY

154 STOICHIOMETRY [CHAP. 10

LIMITING QUANTITIES
10.30. How many sandwiches, each containing 1 slice of salami and 2 slices of bread, can you make with
42 slices of bread and 25 slices of salami?
Ans. With 42 slices of bread, the maximum number of sandwiches you can make is 21. The bread is the limiting
quantity.

10.31. How can you recognize a limiting-quantities problem?
Ans. The quantities of two reactants (or products) are given in the problem. They might be stated in any terms—
moles, mass, etc.—but they must be given for the problem to be a limiting-quantities problem.

10.32. How much sulfur dioxide is produced by the reaction of 5.00 g S and all the oxygen in the atmosphere
of the earth?
Ans. In this problem, it is obvious that the oxygen in the entire earth’s atmosphere is in excess, so that no
preliminary calculation need be done.
S + O 2 −→ SO 2

1 mol S 1 mol SO 2 64.1gSO
5.00gS 2 = 10.0gSO 2
32.06gS 1 mol S 1 mol SO 2
10.33. (a) The price of pistachio nuts is $5.00 per pound. If a grocer has 17 lb for sale and a buyer has $45.00
to buy nuts with, what is the maximum number of pounds that can be sold? (b) Consider the reaction

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

If 45.0 mol of FeCl 2 and 17.0 mol of KMnO 4 are mixed with excess HCl, how many moles of MnCl 2
can be formed?
Ans. (a) With $45, the buyer can buy

1lb
45 dollars = 9.0lb
5 dollars
Since the seller has more nuts than that, the money is in limiting quantity and controls the amount of
the sale.
(b) With 45.0 mol FeCl 2 ,

1 mol KMnO 4
45.0 mol FeCl 2 = 9.00 mol KMnO 4 required
5 mol FeCl 2
Since the number of moles of KMnO 4 present (17.0 mol) exceeds that number, the limiting quantity is
the number of moles of FeCl 2 .

1 mol MnCl 2
45.0 mol FeCl 2 = 9.00 mol MnCl 2
5 mol FeCl 2
10.34. In each of the following cases, determine which reactant is present in excess, and tell how many moles
in excess it is.
Equation Moles Present
(a)2 Na + Cl 2 −→ 2 NaCl 1.20 mol Na, 0.400 mol Cl 2
0.25 mol P 4 O 10 , 1.5 mol H 2 O
(b)P 4 O 10 + 6H 2 O −→ 4H 3 PO 4
(c) HNO 3 + NaOH −→ NaNO 3 + H 2 O 0.90 mol acid, 0.85 mol base
(d) Ca(HCO 3 ) 2 + 2 HCl −→ CaCl 2 + 2CO 2 + 2H 2 O 2.5 mol HCl, 1.0 mol Ca(HCO 3 ) 2
(e) H 3 PO 4 + 3 NaOH −→ Na 3 PO 4 + 3H 2 O 0.70 mol acid, 2.2 mol NaOH