Question

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

Answer 1

$\frac{1}{2}{O}_2\left(g\right)+H_2\left(g\right)\rightleftharpoons H_{2}O\left(g\right)$

As $1$ mole of $H_2\left(g\right)$ reacts with $1/2$ mole of $O_2\left(g\right)$

$\therefore 3$ moles of $H_2\left(g\right)$ will react with= $\frac{1}{2}\times 3$ moles of $O_2\left(g\right)$

So, $H_2$ is limiting reagent.

Now, $1$ mole of $H_2\left(g\right)$ gives $1$ mole of $H_{2}O\left(g\right)$

$\therefore$ $3$ moles of $H_2\left(g\right)$ gives $3$ moles of $H_{2}O\left(g\right)$

Moles $\frac{1}{2}{O}_2\left(g\right)+H_2\left(g\right)⟶H_{2}O\left(g\right)$

at $t=0$ $3$ $3$ -

After reaction $3-\frac{3}{2}=1.5$ $3-3=0$ $3$

Total number of moles of gas present at the end of reaction is

$\Rightarrow n=n_{O_2}+n_{H_{2}O}$

$\Rightarrow n=1.5+3=4.5$ moles of gas.