# When a powdered zinc is heated with sulfur, a violent reaction occurs and zinc sulfide forms: Zn + S_8 ----> Zinc Sulfide Some of the reactants......also combine with oxygen in air to form zinc...

When a powdered zinc is heated with sulfur, a violent reaction occurs and zinc sulfide forms: Zn + S_8 ----> Zinc Sulfide

Some of the reactants...

...also combine with oxygen in air to form zinc oxide and sulfur dioxide. When 95.2 g of Zn reacts with 70.0g of S_8, 105.4 g of Zinc sulfide forms. What is the percent yield of Zinc sulfide? If all the remaining reactants combine with oxygen, how many grams of each of the two oxides form?

Show complete step by step solution to explain the process and answer.

jerichorayel | College Teacher | (Level 2) Senior Educator

Posted on

first we need to get the limiting reagent we do calculations from both Zn and S8 and get the amount of ZnS formed from the reaction

so the balanced reaction would be:

8Zn + S8 = 8ZnS

95.2 gZnx   1mole Zn    x 8 moles ZnS  x  97.445 g ZnS                         .                ---------        -----------       -------------- =141.8899        .                  65.38g Zn     8 moles Zn          1 mole Zn      g ZnS

70.0 g S8 x  1mole S8    x  8 moles ZnS x 97.445 g ZnS                   .                 -----------     --------------   -------------- = 272.728      .                  256.52g S8       1 mole S8       1 mole ZnS      g ZnS

so looking at that we use the mass of ZnS derived from the limiting reagent which is Zn

so we get the actual yield = actual/theoretical  x 100

141.8899 is our theoretical yield. the actual yield is in the problem which is 105.4

so.....    105.4/141.8899   x 100 = 74.28 = 74.3%

no there is an excess which means that we are going to work backwards.. we start from the actual yield of 105.4

105.4 gZnS x 1mole ZnS x 8 moles Zn  x  65.38 g Zn                         .                    ---------     -----------        ---------    =70.7173       .                   97.445 ZnS  8 moles Zns     1 mole Zn      g Zn

remaining Zn = 95.2 - 70.7173 = 24.4827 g Zn

105.4 gZnS x 1mole ZnS x 1mole S8  x  256.52 g S8                         .                    ---------     -----------        ---------    =34.6826       .                   97.445 ZnS  8 moles Zns     1 mole S8      g Zn

remaining S8 = 70.0 - 34.6826 = 35.3174 g S8

the remaining reacts with  Oxygen

2Zn + O2 = 2ZnO

24.4827 g Zn x 1 mole Zn  x 2 moles ZnO x 81.3794 g ZnO           .                      ----------    ---------------  --------------             .                      65.38 g Zn    2 moles Zn         1 mole ZnO

= 30.4739 = 30.5 g ZnO

S8 + 8O2 = 8SO2

35.3174 g S8 x 1 mole S8  x 8 moles SO2 x 64.0644 g SO2           .                      ----------    ---------------  --------------             .                     256.52 g S8    1 mole S8         1 mole SO2

= 70.5625 = 70.6 g SO2

Hope this helps

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