Answer:
To determine the percent yield, we need to compare the actual yield of the reaction to the theoretical yield.
First, we need to balance the chemical equation for the reaction between zinc sulfide and oxygen:
ZnS + O2 → ZnO + SO2
Next, we need to calculate the theoretical yield of zinc oxide based on the amount of limiting reactant (the reactant that is completely consumed in the reaction). To do this, we need to determine which reactant is limiting. We can do this by calculating the amount of product that would be produced from each reactant, and then comparing those amounts to see which one produces less product.
The balanced equation tells us that 1 mole of ZnS reacts with 1 mole of O2 to produce 1 mole of ZnO and 1 mole of SO2. So we can use the molar masses of ZnS and O2 to convert the given masses to moles:
44.5 g ZnS × (1 mol ZnS/97.47 g ZnS) = 0.457 mol ZnS
13.3 g O2 × (1 mol O2/32 g O2) = 0.416 mol O2
According to the balanced equation, 1 mole of ZnS reacts with 1 mole of O2 to produce 1 mole of ZnO. So the limiting reactant is O2, since we have less O2 than we need to react with all of the ZnS. Therefore, the amount of ZnO that can be produced is limited by the amount of O2 available.
The theoretical yield of ZnO can be calculated from the amount of O2:
0.416 mol O2 × (1 mol ZnO/1 mol O2) × (81.39 g ZnO/1 mol ZnO) = 33.7 g ZnO
Now we can calculate the percent yield:
percent yield = (actual yield/theoretical yield) × 100%
We are given that 18.4 g of ZnO is recovered, so the actual yield is 18.4 g.
percent yield = (18.4 g/33.7 g) × 100% ≈ 54.6%
Therefore, the percent yield for the reaction is approximately 54.6%.