Question

When 150. g zinc sulfide are burned in excess oxygen, 68.5 g of zinc oxide are actually produced, along with sulfur dioxide. Determine the percent yield of zinc oxide after finding the theoretical production of zinc oxide.

**Show the balanced equation before you begin.**

Answer 1

%yield = 54.6%

Further explanation

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

(theoretical)

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

Reaction

2ZnS+3O₂ ⇒ 2ZnO+2SO₂

MW ZnS = 97.474 g/mol

• mol ZnS

`\tt (150)/(97.474)=1.54`

MW ZnO = 81.38 g/mol

• mol ZnO (from mol ZnS as limiting reactant, O₂ excess)

`\tt (2)/(2)* 1.54=1.54`

• Actual ZnO produced

`\tt 1.54* 81.38=125.33~g`

Theoretical production = 125.388

• %yield

`\tt (68.5)/(125.33)* 100\%=\boxed{\bold{54.6\%}}`