This is a stoichiometry problem, and as with all stoich problems, we start with our balanced chemical equation. Luckily, it is provided for us:
"2HCl(aq)"+"NaCO"_3(aq)->"2NaCl(aq)"+"H"_2"O"(l)+"CO"_2"(g)"2HCl(aq)+NaCO3(aq)→2NaCl(aq)+H2O(l)+CO2(g)
We are given molarity and volume of sodium carbonate ("Na"_2"CO"_3Na2CO3); we can use this information to find moles of sodium carbonate because moles, volume, and molarity are related by this equation:
"Molarity"="moles of substance"/"liters of substance"Molarity=moles of substanceliters of substance
We have 0.75L0.75L (volume) and 0.3M0.3M (concentration), so plugging into the above equation and solving for moles gives us:
0.3M="moles of substance"/(0.75L)0.3M=moles of substance0.75L
0.225="moles of substance"0.225=moles of substance
So we have 0.225"mol"0.225mol of sodium carbonate. Why do we need this information? Well, now we can calculate the number of moles of hydrochloric acid ("HCl"HCl) from the balanced chemical equation. "HCl"HCl and "Na"_2"CO"_3Na2CO3 react in a 2:1 ratio, because in the equation there's a two in front of "HCl"HCl and an imaginary 1 in front of "Na"_2"CO"_3Na2CO3:
"2HCl"/("1Na"_2"CO"_3)2HCl1Na2CO3
This means, loosely speaking, it takes 2 molecules of "HCl"HCl to combine with 1 molecule of "Na"_2"CO"_3"Na2CO3.
In order to react with 0.225"mol"0.225mol of sodium carbonate, we need:
"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 "HCl", we can use this equation again:
"Molarity"="moles of substance"/"liters of substance"
Filling in and solving produces the final answer:
2.50M="0.45mol HCl"/"liters of substance"
"liters of substance"="0.45mol HCl"/(2.50M)=0.18
So we need "0.180L" of "HCl" to completely react.
