Zinc sulfide reacts with oxygen according to the reaction: 2ZnS (s) 3O2 (g) →2ZnO (s) 2SO2 (g) A reaction mixture initially contains 3.0 mol ZnS and 5.2 mol O2. Part A Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant is left

There will remain 0.7 moles of O2 in excess

Explanation:

Step 1: The balanced equation

2ZnS (s) + 3O2 (g) → 2ZnO (s) + 2SO2 (g)

Step 2: Given data

Number of moles of ZnS = 3.0 mol

Number of moles of O2 = 5.2 mol

Molar mass ZnS = 97.47 g/mol

Molar mass of O2 = 32 g/mol

Step 3: Calculate limiting reactant

For 2 moles of ZnS consumed, we need 3 moles of O2 to produce 2 moles of ZnO and 2 moles of SO2

ZnS is the limiting reactant. It will completely be consumed. (3.0 moles). O2 is the reactant in excess. There will be consumed 3.0 mol * 3/2 = 4.5 mol

There will remain 5.2 - 4.5 = 0.7 moles of O2

There will be produced 3 moles of ZnO and 3 moles of SO2