State the reaction when zinc sulphide reacts with oxygen to produce sulphur dioxide and zinc oxide

Question

a. Determine the mass of zinc oxide that would be produced if 16.7 gram of zinc sulphide is combined with 6.70 gram of oxygen gas 

b. If 7.6 gram are produced determine the percent yeild of reaction 

c. How much of excess reagent is left over at end of reaction 

Answer 1

Write a correctly balanced equation:

2 ZnS(s) + 3 O₂(g) ==> 2 SO₂(g) + 2 ZnO

(a) Find limiting reactant:

16.7 g ZnS x 1 mole/97.4 g = 0.171 mole ZnS

6.7 g O₂ x 1 mole/32 g  = 0.209 moles O₂  <==LIMITING REACTANT

moles ZnO formed = 0.209 moles O2 x 2 moles ZnO/3 moles O2 = 0.139 moles ZnO

mass ZnO = 0.139 moles x 81.4 g/mole = 11.3 grams

 

(b) percent yield = actual/theoretical (x100) = 7.6 g/11.3 g (x100%) = 67.3% yield

 

(c) 0.209 moles O2 x 2 moles ZnS/3 moles O2 = 0.139 moles ZnS used

moles ZnS left over = 0.171 moles - 0.139 moles = 0.032 moles

If you want mass left over, multiply moles x molar mass