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Problem: Zinc sulfide reacts with oxygen according to the reaction: 2 ZnS(s) + 3 O2(g) → 2 ZnO(s) + 2 SO2(g) A reaction mixture initially contains 4.2 mol ZnS and 8.7 mol O2.Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant remains?

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We are asked to calculate the amount (in moles) of the excess reactant. 


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


Calculate the mole ZnO based on both reactants: 


4.2 mol ZnS × 2 molZnO2 mol ZnS  = 4.2 mol ZnO8.7 mol O2 × 2 molZnO3 mol O2  = 5.8 mol ZnO

Since ZnS produced less product, it is limiting. 


Calculate O2 used.  

4.2 mol ZnS × 3 molO22 mol ZnS  = 6.3 mol O2


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Problem Details

Zinc sulfide reacts with oxygen according to the reaction:

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

A reaction mixture initially contains 4.2 mol ZnS and 8.7 mol O2.

Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant remains?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Limiting Reagent concept. If you need more Limiting Reagent practice, you can also practice Limiting Reagent practice problems.