Hi The reaction between ZnS and O2 is, 2ZnS (s) + 3O2 (g) --> 2ZnO (s) + 2SO2 (g

Question

Hi

The reaction between ZnS and O2 is,

2ZnS (s) + 3O2 (g) --> 2ZnO (s) + 2SO2 (g)

So from the stoichiometry of the reaction, it is evident that,

2 moles of ZnS react with = 3 moles of O2

So, 0.250 moles of ZnS react with= (3/2)*0.250 moles of O2

= 0.375 moles of O2.

So the required amount of O2 for the given amount of ZnS completely react is higher than the available amount. Therefore, O2 is limiting reactant and ZnS is excess reactant.

Again, from stoichiometry,

0.250 moles of O2 react with = (2/3)*0.250 moles of ZnS

= 0.167 moles

So moles of excess reactant left= Reactant available - Reactant reacted

= 0.250 - 0.167 moles

= 0.083 mol of ZnS. Answer

Thank you

Explanation / Answer

(A) 160 g (B) 176g (C) 639g (D) 9420 g A reaction mixture initially contains 0.250 mol of Zns and 0.250 mol of O2. After the mixture has reacted ompletely, how many moles of the excess reactant are left? 52. cid react 2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g) (A) 0.083 mol ZnS (C) 0.125 mol O2 (B) 0.167 mol ZnS (D) 0.250 mol O2

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