Question

$3.0moles$ of oxygen gas react with $3.0$ moles of hydrogen gas to produce water vapor. What are the total moles of gas present at the end of the reaction?

• A. $1.5moles$ of gas are present
• B. $2.5moles$ of gas are present
• C. $3.0moles$ of gas are present
• D. $4.5moles$ of gas are present

Answer 1

The correct option is D $4.5moles$ of gas are present

The reaction is given by,

$2H_2+O_2⟶2H_{2}O$

i.e. $2\text{mol}$ of $H_2+1\text{mole}$ of $O_2⟶2\text{mole}$ of $H_{2}O$

Or,

$6\text{mol}$ of $H_2+3\text{mole}$ of $O_2⟶6\text{mole}$ of $H_{2}O$

Given : $3\text{mol}$ of $H_2+3\text{mole}$ of $O_2$ react,

$⟹3\text{mol}$ of $H_2+1.5\text{mole}$ of $O_2⟶3\text{mole}$ of $H_{2}O$.

So, $(3-1.5)=1.5\text{mol}$ of $O_2$ is unreacted, and $3\text{mol}$ of $H_{2}O$ vapor is produced.

$\therefore$ Total moles of gas present at the end of the reaction$=3+1.5=4.5\text{moles}.$