Question

How many grams of $O_2$ will it take to oxidize $88grams$ 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 $224$

The reaction for the oxidation of Propane can be written as-

$2C_3{H}_8$ $+$ $7O_2$ $\to$ $6CO$ $+$ $8H_{2}O$

Molecular weight of Propane is $44grams$. Thus two moles of Propane weighs $88grams$.

The molecular weight of Oxygen is $32grams$. Thus $7moles$ of Oxygen weighs ($7$$\times$$32$)grams, that is $224grams$

According to the equation,

$88grams$ of Propane require ($7$$\times$$32$)$grams$ of Oxygen that is $224grams$ of Oxygen to be oxidized to Carbon monoxide and Water.

Thus, option D is the correct answer.