Answer
11.0 g ZnS was produced.
Work Step by Step
Strategy: 1- Find the moles of ZnS each given amount of reactant would give. The one that gives the smallest amount is the limiting reactant.
2- Convert the moles of ZnS to gram
Step 1: 8 Zn + $S_{8}$ ${\longrightarrow}$ 8 ZnS
find how many moles of ZnS can be obtained from 7.36 g Zn using the converting factors:
7.36 g Zn x $\frac{1 mol(Zn)}{Molar Mass (Zn)}$ x $\frac{8 mol (ZnS)}{8 mol(Zn)}$ =
7.36 g Zn x $\frac{1 mol(Zn)}{65.4 g (Zn)}$ x $\frac{8 mol (ZnS)}{8 mol(Zn)}$ = 0.113 mol ZnS
6.45 g $S_{8}$ x $\frac{1 mol(S_{8})}{Molar Mass (S_{8})}$ x $\frac{8 mol (ZnS)}{1 mol(S_{8})}$ =
6.45 g $S_{8}$ x $\frac{1 mol(S_{8})}{256g (S_{8})}$ x $\frac{8 mol (ZnS)}{1 mol(S_{8})}$ = 0.202 mol ZnS
7.36 g Zn gives the smaller amount of ZnS, so Zn is the limiting reactant.
Step 2: Convert the 0.113 moles of ZnS that can be obtained from 7.36 g Zn, to gram ZnS:
0.113 mol ZnS x $\frac{Molar Mass(ZnS)}{1 mol (ZnS)}$=
0.113 mol ZnS x $\frac{97.4 g(ZnS)}{1 mol (ZnS)}$= 11.0 g ZnS
