Question

How many grams of $O_2$ will require to oxidize 88 grams of $C_3{H}_8$ to $CO$ and $H_{2}O$?

• A. 32
• B. 64
• C. 112
• D. 166
• E. 224

Answer 1

Answer: (E)

224

$C_3{H}_8+\frac{7}{2}{O}_2\to 3CO+4H_{2}O$
Molar mass of $C_3{H}_8$ = 44 g/mol.
In 88 g of $C_3{H}_8$ there are 2 moles of $C_3{H}_8$.
As per reaction for 1 mol of $C_3{H}_8$, there is required $\frac{7}{2}$ moles of $O_2$ = 112 g.
So, for 2 moles of $C_3{H}_8$, $112\times 2=224$ g of $O_2$ is required.