When 150.0 g of ZnS are burned in excess oxygen, 68.5 g of ZnO are produced according to the equation:
2ZnS + 302 - 2ZnO + 2SO2
What is the percent yield for this reaction?

Question

When 150.0 g of ZnS are burned in excess oxygen, 68.5 g of ZnO are produced according to the equation:
2ZnS + 302 -> 2ZnO + 2SO2
What is the percent yield for this reaction?

jfraser · Answer

Use stoichimetry to find out how much ZnO "should have been" made, and compare that to the amount that was actually made.$$150.0g Zn * (\frac{1 mol Zn}{MM Zn})*(\frac{mol ZnO}{mol Zn})*(\frac{MMZnO}{mol ZnO})$$ where MM is the molar mass of the stuff in that space. When you find an amount this way, you are finding a theoretical yield.

Percent yield is$$%Y = \frac{actual yield}{theoretical yield}$$

anonymous · Answer

%yield is the ratio of actual yield/theoretical yield.