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1. when ZnS + o2 → ZnO + SO2rnrn93.4 grams of o2 are reacted with 471.51 grams of ZnS, 1.63 moles of ZnO are produced. What is the percent yield of this reaction?rn% yield
Accepted Answer
1) ZnS + O2 → ZnO + SO2
93.4 grams of O2 are reacted with 471.51 grams of ZnS, 1.63 moles of ZnO are produced. What is the percent yield of this reaction?
Percent Yield
The percent yield of a reaction is a measure of the efficiency of the reaction. It is calculated by dividing the actual yield of the reaction by the theoretical yield and multiplying by 100%.
The actual yield of the reaction is the amount of product that is actually obtained from the reaction. In this case, the actual yield is 1.63 moles of ZnO.
The theoretical yield of the reaction is the amount of product that would be obtained if the reaction went to completion. In this case, we can use the balanced chemical equation to calculate the theoretical yield.
1 mole ZnS + 1 mole O2 → 1 mole ZnO + 1 mole SO2
From this equation, we can see that 1 mole of ZnS will react with 1 mole of O2 to produce 1 mole of ZnO. Therefore, the theoretical yield of ZnO is 471.51 g / 97.44 g/mol = 4.84 moles.
The percent yield of the reaction is calculated as follows:
% yield = (actual yield / theoretical yield) * 100%
% yield = (1.63 mol / 4.84 mol) * 100%
= 33.7%
Therefore, the percent yield of this reaction is 33.7%.
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