# Question a69d3

Nov 11, 2016

$\text{0.180L HCl}$

#### Explanation:

This is a stoichiometry problem, and as with all stoich problems, we start with our balanced chemical equation. Luckily, it is provided for us:
$\text{2HCl(aq)"+"NaCO"_3(aq)->"2NaCl(aq)"+"H"_2"O"(l)+"CO"_2"(g)}$

We are given molarity and volume of sodium carbonate (${\text{Na"_2"CO}}_{3}$); we can use this information to find moles of sodium carbonate because moles, volume, and molarity are related by this equation:
$\text{Molarity"="moles of substance"/"liters of substance}$

We have $0.75 L$ (volume) and $0.3 M$ (concentration), so plugging into the above equation and solving for moles gives us:
$0.3 M = \frac{\text{moles of substance}}{0.75 L}$
$0.225 = \text{moles of substance}$

So we have $0.225 \text{mol}$ of sodium carbonate. Why do we need this information? Well, now we can calculate the number of moles of hydrochloric acid ($\text{HCl}$) from the balanced chemical equation. $\text{HCl}$ and ${\text{Na"_2"CO}}_{3}$ react in a 2:1 ratio, because in the equation there's a two in front of $\text{HCl}$ and an imaginary 1 in front of ${\text{Na"_2"CO}}_{3}$:
"2HCl"/("1Na"_2"CO"_3)#

This means, loosely speaking, it takes 2 molecules of $\text{HCl}$ to combine with 1 molecule of $\text{Na"_2"CO"_3}$.

In order to react with $0.225 \text{mol}$ of sodium carbonate, we need:
$\text{2HCl"/cancel("1Na"_2"CO"_3)xx0.225"mol"cancel(" Na"_2"CO"_3)=0.45"mol HCl}$

The problem is asking us for volume, though, and since we know molarity and moles of $\text{HCl}$, we can use this equation again:
$\text{Molarity"="moles of substance"/"liters of substance}$

Filling in and solving produces the final answer:
$2.50 M = \text{0.45mol HCl"/"liters of substance}$
$\frac{\text{liters of substance"="0.45mol HCl}}{2.50 M} = 0.18$

So we need $\text{0.180L}$ of $\text{HCl}$ to completely react.