Zns (zinc sulfide) (Zn = 65.38; S = 32.06] *

The question is incomplete, the complete question is;

Zinc and sulfur react to form zinc sulfide by the equation shown below. How many grams of ZnS can be formed when 12.0 g of Zn reacts with 6.50 g of S? (Atomic mass: Zn = 65.38, S = 32.06).

17.5 g of ZNS

Explanation:

The equation of the reaction is;

Zn(s) + S(s) >ZnS(s)

Number of moles of Zn = 12.0/65.38 = 0.18 moles

Number of moles of S = 6.50/32.06 = 0.20 moles

Hence Zn is the limiting reactant

If 1 mole of Zn yields 1 mole of ZnS

Then 0.18 moles of Zn also yields 0.18 moles of ZnS

Mass of ZnS produced = 0.18 moles *  97.44 g/mol = 17.5 g of ZNS

u i forgot the answer?

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