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QUESTION

# What is the limiting reagent between a reaction of HCl(aq) and NaOH(aq)?

There is no here.

"HCl"_text((aq]) + "NaOH"_text((aq]) -> "NaCl"_text((aq]) + "H"_2"O"_text((l])

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

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

color(blue)(c = n/V implies n = c * 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 , 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_(HCl) = "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 "1-M" sodium hydroxide solution would contain that many moles of sodium hydroxide?

c = n/V implies V = 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.