Stoichiometry ‹ 93
Answer:- Convert to moles:
()
=
=
125 g P O1 molPO
142 g PO0.880molPO(50.0 gHO)
1 molHO
18.0 gHO2.78 mol HO2525
25252
2
22- Find the limiting reactant:
0.880 mol
1 mol
0.880 P O
2.78 mol
3 mol0.927 H O
252=
=
The 1 mol and the 3 mol come from the balanced chemical equation. The 0.880 is smaller,
so this is the L.R.- Finish using the number of moles of the L.R.:
()
0.880molPO =
2 molHPO
1 mol PO98.0 g HPO
1 mol HPO
25 34 172 g
2534
34Percent Yield
In the preceding problems, the amount of product calculated based on the limiting-
reactant concept is the maximum amount of product that could be formed from the given
amount of reactants. This maximum amount of product formed is called the theoretical
yield. However, rarely is the amount that is actually formed (the actual yield) the same as
the theoretical yield. Normally it is less. There are many reasons for this, but the principal
reason is that most reactions do not go to completion; they establish an equilibrium system
(see Chapter 15, Equilibrium, for a discussion of chemical equilibrium). For whatever
reason, not as much as expected is formed. The efficiency of the reaction can be judged by
calculating the percent yield. The percent yield (% yield) is the actual yield divided by the
theoretical yield, and the result is multiplied by 100% to generate percentage:% yield=×actualyield
theoreticalyield100%
Consider the problem in which it was calculated that 60.8 g NH 3 could be formed. Suppose
that reaction was carried out, and only 52.3 g NH 3 was formed. What is the percent yield?% yield=×=52.3 g
60.8 g100% 86.0%
Let’s consider another percent yield problem in which a 25.0-g sample of calcium oxide
is heated with excess hydrogen chloride to produce water and 37.5 g of calcium chloride.
What is the percent yield of calcium chloride?
CaO(s) + 2HCl(g) → CaCl 2 (aq) + H 2 O(l)
Answer:
(25.0 gCaO) =
1mol CaO
56.077gCaO1mol CaCl
1mol CaO110.984gCaCl
1mol CaCl(^22) 49.478gCaCl
2
2