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)"#

We are given molarity and volume of sodium carbonate (#"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:

#"Molarity"="moles of substance"/"liters of substance"#

We have #0.75L# (volume) and #0.3M# (concentration), so plugging into the above equation and solving for moles gives us:

#0.3M="moles of substance"/(0.75L)#

#0.225="moles of substance"#

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

#"2HCl"/("1Na"_2"CO"_3)#

This means, loosely speaking, it takes 2 molecules of #"HCl"# to combine with 1 molecule of #"Na"_2"CO"_3"#.

In order to react with #0.225"mol"# 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.