# Question #421e4

Aug 18, 2017

True.

#### Explanation:

The key here is the balanced chemical equation that describes the formation of water from hydrogen gas and oxygen gas

$2 {\text{H"_ (2(g)) + "O"_ (2(g)) -> 2"H"_ 2"O}}_{\left(l\right)}$

Two things to notice here

$\textcolor{b l u e}{\left(1\right)}$. The number of moles of water produced by the reaction is equal to the number of moles of hydrogen gas consumed by the reaction $\to$ the two chemical species are in a $2 : 2$ mole ratio.

$\textcolor{b l u e}{\left(2\right)}$. The reaction consumes $2$ moles of hydrogen gas for very $1$ mole of oxygen gas that takes part in the reaction $\to$ the two reactants are in a $2 : 1$ mole ratio.

In your case, you know that you are mixing $2$ moles of hydrogen gas and $3$ moles of oxygen gas.

According to $\textcolor{b l u e}{\left(2\right)}$, $2$ moles of hydrogen gas will consume $1$ mole of oxygen gas, so right from the start, you can say that oxygen gas is in excess.

In other words, hydrogen gas will act as the limiting reagent, i.e. it will be completely consumed before all the moles of oxygen gas will get the chance to react.

So if the reaction consumes $2$ moles of hydrogen gas and $1$ mole of oxygen gas, it follows that it will produce $2$ moles of water $\to$ this is what $\textcolor{b l u e}{\left(1\right)}$ tells you.

Therefore, you can say that the affirmation is true--when $2$ moles of hydrogen react with $3$ moles of oxygen, $2$ moles of water are produced.

You can add to this by saying that $2$ moles of oxygen gas are left unreacted--remember, the reaction will only consume $1$ mole of oxygen gas before all the moles of hydrogen gas will be completely consumed.