Answer:1. To determine the limiting reactant, we need to calculate the amount of ZnO and SO2 that can be produced from each reactant.
From the equation, 2 moles of ZnS reacts with 3 moles of O2 to produce 2 moles of ZnO and 2 moles of SO2.
First, convert the masses of ZnS and O2 to moles:
19.6 g of ZnS = 19.6 g / 97.4 g/mol = 0.200 mol ZnS
33.6 g of O2 = 33.6 g / 32.0 g/mol = 1.05 mol O2
Next, calculate the moles of ZnO and SO2 that can be produced by the limiting reactant:
0.200 mol ZnS * 2 moles ZnO / 2 moles ZnS = 0.200 mol ZnO
0.200 mol ZnS * 2 moles SO2 / 2 moles ZnS = 0.200 mol SO2
1.05 mol O2 * 2 moles ZnO / 3 moles O2 = 0.700 mol ZnO
1.05 mol O2 * 2 moles SO2 / 3 moles O2 = 0.700 mol SO2
Since 0.200 mol ZnO and 0.200 mol SO2 are less than 0.700 mol ZnO and 0.700 mol SO2 produced by O2, ZnS is the limiting reactant.
2. To find the amount of SO2 produced, simply use the amount of ZnS as a limiting reactant:
0.200 mol ZnS * 2 moles SO2 / 2 moles ZnS = 0.200 mol SO2
3. To find the amount of SO2 produced in grams, multiply the moles of SO2 by its molar mass:
0.200 mol SO2 * 64.1 g/mol = 12.8 g SO2
3. To find the amount of SO2 produced in grams, multiply the moles of SO2 by its molar mass:
0.200 mol SO2 * 64.1 g/mol = 12.8 g SO2
4. The excess reactant is O2, since there is 1.05 mol of O2 and only 0.200 mol ZnS is needed to react with it. The remaining 0.850 mol O2 is the excess reactant.
Step-by-step explanation: