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Agata [3.3K]
2 years ago
14

Zinc sulfide reacts with oxygen according to the reaction 2ZnS(s)+3O2(g)→2ZnO(s)+2SO2(g) A reaction mixture initially contains 5

.0 molZnS and 9.8 molO2. Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant remains? Express your answer using two significant figures.
Chemistry
1 answer:
LenKa [72]2 years ago
4 0

Answer:

The answer to your question is: Excess oxygen = 2.3 mol

Explanation:

Data

ZnS = 5 mol

O2 = 9.8 mol

Excess reactant = ?

Balanced reaction

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

MW ZnS = 65 + 32 = 97 x 2 = 194 g

MW O2 = 16 x  6 = 96 g

                           2 mol of ZnS  ------------------- 3 mol O2

 

Ratio from the reaction = 3 mol O2/ 2 mol ZnS

                                      = 1.5

Ratio from the quantities in the experiment = 9.8 mol O2 / 5 mol of ZnS

                                                                        = 1.96

Excess reactant = Oxygen because the ratio increases

                             

                                2 mol of ZnS  ------------------- 3 mol O2

                                5 mol of ZnS  -------------------  x

                                x = (5 x 3) / 2

                               x = 7.5 mol of O2

Excess Oxygen = 9.8 mol - 7.5 mol

Excess oxygen = 2.3 mol

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