Zinc(II)sulfide reacts with oxygen: 2ZnS(s)+3O2(g)--->2ZnO(s)+2SO2(g) A reaction mixture initially contains 4.6 mol ZnS and 7.2 mol O2. Once the reaction is completely as possible, what amount (in moles) of the excess reactant is left?


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John McCann Profile
John McCann answered
I suspect ZnS limits. Let's see.

7.2 mole O2 (2 mole ZnS/3 mole O2) = 4.8 mole ZnS

you only have 4.6 mole ZnS, so it limits

4.6 mole ZnS (3 mole O2/2 mole ZnS) = 6.9 mole O2

you have more O2 than 6.9 mole (7.2 mole O2 ), so you used up 6.9 mole O2.

I think you can figure out how many moles of the excess reactant are left.

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