Considering the limiting reactant concept,how many moles of lead(IV)sulfide are produced from the reaction of 0.500 mol of lead and 0.750 mol of sulfur? Pb(s)+ 2 S(s) PbS₂(s)

Question

Considering the limiting reactant concept,how many moles of lead(IV)sulfide are produced from the reaction of 0.500 mol of lead and 0.750 mol of sulfur? Pb(s)+ 2 S(s)  PbS₂(s)

Quizplus · Accepted Answer

First, we need to determine the limiting reactant by using stoichiometry. The balanced equation is:

Pb (s) + 2S (s) → PbS₂ (s)

The mole ratio between lead and lead(IV) sulfide is 1:1, and the mole ratio between sulfur and lead(IV) sulfide is 2:1. Let's calculate the amount of product that can be formed from each reactant:

From lead: 0.500 mol Pb × (1 mol PbS₂ / 1 mol Pb) = 0.500 mol PbS₂

From sulfur: 0.750 mol S × (1 mol PbS₂ / 2 mol S) = 0.375 mol PbS₂

Since the amount of lead(IV) sulfide produced from sulfur is less than the amount produced from lead, sulfur is the limiting reactant. Therefore, the amount of lead(IV) sulfide produced is 0.375 mol. The best choice is B.

Considering the Limiting Reactant Concept,how Many Moles of Lead(IV)sulfide Are