Calculating heat capacity of a calorimeter

Question

I am struggling on this question :

The heat of neutralization of $\ce{NaOH}$ and $\ce{HCl}$ is $-57.1\:\mathrm{kJ}$, calculate the numbers of moles of $\ce{HCl}$ and if $x$ moles of $\ce{HCl}$ were used how many $\:\mathrm{kJ}$ would be given out if one mole results in $57.1 \:\mathrm{kJ}$

I already got the moles as $0.05$ by doing $n=\frac{v}{1000} c$ where I had $50\:\mathrm{mL}$ of $2\:\mathrm{M}$ $\ce{HCl}$

Answer 1

First, is this a question from your book, and do you have the correct answer to check your work.

Do you know how to use dimensional analysis?

Start with 50 mL. You have 2 Molarity of some subtance, how do you write that as a quotient and what are the units of the denominator and numerator? You will also need to know how many mL are in a liter and you should get the correct moles of that substance. It is very important that you write this down on paper correctly and make sure your units will cancel leaving only moles. It is hard to be wrong when your units cancel correctly.

Do you mean the heat of neutralization is $-51.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}$

The more information you give us, the more help you will get.

Answer 2

First lets work out moles of $\ce{HCl}$. The concentration is $2\ \mathrm M$, or in other words, 2 moles per liter. To figure out how many moles you have then, you need to figure out how many liters you have. $50\ \mathrm{mL} = 5\times10^{-2}\ \mathrm L$.

Multiplying concentration and volume, you are saying $\frac{\mathrm{mol}}{\mathrm L}\cdot\mathrm L=\mathrm{mol}$. So you have $2\ \mathrm M\times5\times10^{-2}\ \mathrm L=0.1\ \mathrm{mol}$

You then return to some information you've been given. The heat of neutralization is: $-51.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}$. So to get the total energy in kJ, you can see by the unit that you have to multiply by the number of moles. That cancels the units like we did with the volume: $$-51.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}\times0.1\ \mathrm{mol}=-5.17\ \mathrm{kJ}$$