What is the percent yield if 56.9 g of WO3 yields 41.4 g of tungsten?

Question

WO3 + H2 → W + H2O

This reaction is carried out in the lab, and 11.4 g of tungsten are obtained. If the percent yield is 89.4%, what mass of WO3 was used?

Answer 1

The balanced equation is
2 WO3 +6 H2 >> 2 W + 6H2O

Molecular weight WO3 = 231.85 g/mol
56.9 g / 231.85 = 0.245 moles WO3

atomic weight W = 183.85 g/mol
41.4 g / 183.85 = 0.225 moles W
The ratio between WO3 and W is 2 : 2 or 1 : 1
we would get 0.245 moles W
0.225 : 0.245 = x : 100
x = 91.8%

About the 2nd question

11.4 /183.85 = 0.0621 moles obtained


100 : 89.4 = x : 0.0621

x = 0.0695 moles WO3 used
0.0695 mol x 231.85 = 16.1 g WO3

Answer 2

Here is another way to solve:

determine weight fraction of tungsten in WO3:

183.85/[183.85 + [3 x 16]] = 0.793

then, use to calculate the theoretical yield of W:

56.9g x 0.793 = 45.12g

then determine percent yield:

41.4g/[56.9g x 0.793] = 91.8%

For the second part:

11.4g x 1/0.894 x 1/0.793 = 16.08g

Where the term 1/0.894 corrects for the less than 100% yield (89.4%), and the 1/0.793 term gives the corresponding weight of WO3 for a given weight of tungten.