Answer
40.32 g C$H_{3}$OH
The amount of unconsumed $H_{2}$ : 10.2 g - 5 g = 5.2 g
Work Step by Step
Strategy: 1- Find the moles of C$H_{3}$OH that each given amount of reactant would give. The one that gives the smallest amount is the limiting reactant.
2- Convert the moles of C$H_{3}$OH to mass, in gram
3- Find the amount of unconsumed reactant
Step 1: CO + 2 $H_{2}$ ${\longrightarrow}$ C$H_{3}$OH
find how many moles of C$H_{3}$OHcan be obtained from 35.4 CO using the converting factors:
35.4 g CO x $\frac{1 mol(CO)}{Molar Mass (CO)}$ x $\frac{1 mol (CH_{3}OH)}{1 mol(CO)}$ =
35.4 g COx $\frac{1 mol(CO)}{28 g (CO)}$ x $\frac{1 mol (CH_{3}OH)}{1 mol(CO)}$ = 1.26 mol C$H_{3}$OH
10.2 g $H_{2}$ x $\frac{1 mol(H_{2})}{Molar Mass (H_{2})}$ x $\frac{1 mol (CH_{3}OH)}{2 mol(H_{2})}$ =
10.2 g $H_{2}$ x $\frac{1 mol(H_{2})}{2 g (H_{2})}$ x $\frac{1 mol (CH_{3}OH)}{2 mol(H_{2})}$ = 2.55 mol C$H_{3}$OH
35.4 g CO gives the smaller amount of C$H_{3}$OH, so COis the limiting reactant.
Step 2: Convert the 1.26 mol C$H_{3}$OH to gram:
1.26 mol C$H_{3}$OH x $\frac{Molar Mass (CH_{3}OH)}{1 mol (CH_{3}OH)}$ =
1.26 mol C$H_{3}$OH x $\frac{32g (CH_{3}OH)}{1 mol (CH_{3}OH)}$ =40.32 g C$H_{3}$OH
3- Convert the mol of C$H_{3}$OH to gram $H_{2}$, to find how many grams $H_{2}$ were consumed in reaction:
1.26 mol C$H_{3}$OH x $\frac{2 mol (H_{2})}{1 mol(CH_{3}OH)}$ x $\frac{Molar mass(H_{2})}{1 mol (H_{2})}$=
1.26 mol C$H_{3}$OH x $\frac{2 mol (H_{2})}{1 mol(CH_{3}OH)}$ x $\frac{2 g(H_{2})}{1 mol (H_{2})}$= 5 g $H_{2}$
The amount of unconsumed $H_{2}$ : 10.2 g - 5 g = 5.2 g
