# Question 10285

Feb 24, 2016

There is no limiting reagent here.

#### Explanation:

${\text{HCl"_text((aq]) + "NaOH"_text((aq]) -> "NaCl"_text((aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

You have a $1 : 1$ mole ratio between hydrochloric acid and sodium hydroxide, which means that a complete neutralization requires equal number of moles of each reactant.

Now, the number of moles of each reactant can be found by using the molarities and volumes of the two solutions.

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

Even without doing any calculations, it should be clear that because you have equal molarities and equal volumes, you will also have equal numbers of moles of hydrochloric acid and sodium hydroxide.

This of course implies that you're not dealing with a limiting reagent, since you have enough strong base to completely neutralize the strong acid (or vice versa).

So unless the values given to you are incorrect, the answer to this question is that neither the hydrochloric acid, nor the sodium hydroxide acts as a limiting reagent here.

If you want to do some unnecessary math, you can try

${n}_{H C l} = \text{1 mol" color(red)(cancel(color(black)("dm"^(-3)))) * 50 * 10^(-3)color(red)(cancel(color(black)("dm"^(3)))) = "0.05 moles HCl}$

What volume of the $\text{1-M}$ sodium hydroxide solution would contain that many moles of sodium hydroxide?

$c = \frac{n}{V} \implies V = \frac{n}{c}$

V_(NaOH) = (0.050color(red)(cancel(color(black)("moles"))))/(1color(red)(cancel(color(black)("mol")))"dm"^(-3)) = "0.050 dm"^3 = "50 cm"^3#

And there you have it, you have enough moles of strong base to completely neutralize the strong acid.