26.4 g of hydrogen, H2, is produced from 100.0 g of methane, CH4, reacting with excess water as shown, what is the percentage yield? CH4 (g) + H2O (g) → CO (g) + 3 H2 (g)

Question

Select one:

a. 

100%

b. 

69.9%

c. 

23.4%

d. 

57.1%

Answer 1

CH4 + H2O ==> CO + 3H2

% yield = actual yield / theoretical yield (x100%)

So we first need to find the theoretical yield

100.0 g CH4 x 1 mol CH4 / 16 g = 6.250 mols CH4

6.250 mol CH4 x 3 mols H2 / mol CH4 x 2 g / mol H2 = 37.5 g H2 … theoretical yield

% yield = 26.4 g / 37.5 g (x100) = 70.4%. ANSWER B (note I rounded off molar masses)