Answer
46.7 g SO2
The unconsumed amount of $CS_{2}$: 35 g - 27.7 g = 7.26 g
Work Step by Step
Strategy: 1- Find the moles of S$O_{2}$ that each given amount of reactant would give. The one that gives the smallest amount is the limiting reactant.
2- Convert the moles of S$O_{2}$ to mass, in gram
3- Find the amount of unconsumed reactant
Step 1: C$S_{2}$ + 3$O_{2}$ ${\longrightarrow}$ C$O_{2}$ + 2S$O_{2}$
find how many moles of S$O_{2}$ can be obtained from 35 g CS2 using the converting factors:
35 g CS2 x $\frac{1 mol(CS2)}{Molar Mass (CS2)}$ x $\frac{2 mol (SO_{2} )}{1 mol(CS2)}$ =
35 g CS2 x $\frac{1 mol(CS2)}{76 g (CS2)}$ x $\frac{2 mol (SO_{2} )}{1 mol(CS2)}$ = 0.92 mol S$O_{2}$
35 g $O_{2}$ x $\frac{1 mol(O_{2})}{Molar Mass (O_{2})}$ x $\frac{2 mol (SO_{2} )}{3 mol(O_{2})}$ =
35 g $O_{2}$ x $\frac{1 mol(O_{2})}{32g (O_{2})}$ x $\frac{2 mol (SO_{2} )}{3 mol(O_{2})}$ = 0.73 mol S$O_{2}$
35 $O_{2}$ gives the smaller amount of S$O_{2}$ , so O2 is the limiting reactant.
Step 2: Convert the 0.73 mol S$O_{2}$ to gram:
0.73 mol S$O_{2}$ x $\frac{Molar Mass (SO2)}{1 mol (SO2)}$ =
0.73 mol S$O_{2}$ x $\frac{64g (SO2)}{1 mol (SO2)}$ = 46.7 g SO2
3- Convert the mol of SO2 to gram C$S_{2}$, to find how many grams C$S_{2}$ were consumed in reaction:
0.73 mol S$O_{2}$ x $\frac{1mol (CS_{2})}{2 mol(SO_{2})}$ x $\frac{Molar mass(CS_{2})}{1 mol (CS_{2})}$=
0.73 mol S$O_{2}$ x $\frac{1mol (CS_{2})}{2 mol(SO_{2})}$ x $\frac{76 g(CS_{2})}{1 mol (CS_{2})}$= 27.7 g $CS_{2}$
The unconsumed amount of $CS_{2}$: 35 g - 27.7 g = 7.26 g
