Question

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

• A. 32
• B. 64
• C. 112
• D. 224

Answer 1

The correct option is D 224

$C_3{H}_8+\frac{7}{2}{O}_2\to 3CO+4H_{2}O$

According to balanced equation: $44g$ $\frac{7}{2}\times 32g$ $3\times 28g$ $4\times 18g$

Since $44g$ of $C_3{H}_8$ is oxidized by $\frac{7}{2}\times 32g$ of $O_2$

So, $88g$ of $C_3{H}_8$ will be oxidized by: $=88\times \frac{7}{2}\times 32\times \frac{1}{44}g$ of $O_2$

$=224g$ of $O_2$