Determine the mass of zinc sulphide, ZnS, which are produced when 6.2 g of zinc and 4.5g of sulphur are reacted. 

Expert Answers
jerichorayel eNotes educator| Certified Educator

In order to solve for the amount of the Zinc sulphide to be produced, we should first establish the reaction equation.

The equation for the reaction of Zinc and Sulphur is:

Zn + S -> ZnS

The next step is to determine the limiting reactant. The limiting reactant will determine the amount of Zinc Sulphide possible.

`4.5 grams S * (1 mol e S)/(32.065grams S) * (1 mol e ZnS)/(1mol e S)`

= 0.1403 moles ZnS


`6.2 grams Zn * (1mol e Zn)/(65.38 grams Zn) * (1 mol e ZnS)/(1mol e Zn)`

= 0.09483moles ZnS 

Since, Zn produces the smaller amount of product, it is the limiting reactant. To get the mass of ZnS, multiply the moles of ZnS with its molecular mass. The molar mass for ZnS is 97.445 grams/mol

`0.09843 mol es ZnS * (97.445 (g)/(mol))`

= 9.24 grams of ZnS

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