# Calculating Average Rate

A $\pu{3.45 g}$ piece of marble ($\ce{CaCO3}$) is weighed and dropped into a beaker containing $\pu{1.00 L}$ of hydrochloric acid. The marble is completely gone $\pu{4.50 min}$ later. Calculate the average rate of reaction of $\ce{HCl}$ in $\pu{mol/L/s}$. Note that the volume of the system remains at $\pu{1.00 L}$ through the entire reaction.

I'm not very sure why the units are in $\pu{mol/L/s}$ instead of just $\pu{mol/s}$. Here is what I did:

$$\text{Rate} = \frac{\pu{3.45 g}~\ce{(CaCO3)}}{\pu{4.5 min}} = \pu{0.0128 g/s}$$

$$\pu{0.128 g}~\ce{CaCO3} = \pu{1.28e-3 mol}$$

$$n(\ce{HCl}) = \pu{2.58e-3 mol}$$

$$\text{Rate} = \pu{2.58e-3 mol/L/s}$$

However, the answer given is $\pu{2.55e-4 mol/L/s}$. What am I doing wrong?

You haven't accounted for the stoichiometry of the reaction, and I suppose you wrongly converted minutes to seconds. Always start solving problems like this with writing down the chemical reaction:

$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$

By definition rate of consumption of hydrochloric acid over time $\Delta t$ is:

$$r=\frac{\Delta c(\ce{HCl})}{\Delta t}$$

Since all calcium carbonate reacted completely:

$$\Delta c(\ce{HCl}) = \frac{\Delta n(\ce{HCl})}{V} = \frac{2n(\ce{CaCO3})}{V} = \frac{2m(\ce{CaCO3})}{V M(\ce{CaCO3})}$$

where $m$ is mass, $M$ - molar mass, $V$ - volume.

And the average rate is:

$$r = \frac{2m(\ce{CaCO3})}{V M(\ce{CaCO3})\Delta t} = \frac{2\cdot\pu{3.45 g}}{\pu{1 L}\cdot\pu{100.09 g mol-1}\cdot\pu{4.50 min}\cdot\pu{60 s min-1}} = \pu{2.55e-4 mol L-1 s-1}$$

Also, be careful with notations. Use proper capitalization, and don't equate moles to grams! This is not tolerable in natural sciences.