# Molarity of silver(I) nitrate solution after the reaction with magnesium [closed]

$\pu{3 g}$ of $\ce{Mg}$ are placed in $\pu{500 mL}$ of $\pu{0.625 M}$ $\ce{AgNO3}$. When the reaction is complete, what is the molarity of the $\ce{AgNO3}$ solution?

I'm thinking you need to use $M_1V_1=M_2V_2$, but I'm stuck on how to use it.

## closed as off-topic by Mithoron, airhuff, John Snow, Jon Custer, TyberiusJan 24 '18 at 20:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

• Try to think of what the definitions of M and V are, what do the subscripts mean? Units are always a good indicator. – John Snow Jan 24 '18 at 6:42

Whenever possible, start with writing down equation(s) for the chemical reactions taking place in the system and don't use random formulas. Magnesium as a more active metal is going to reduce silver in the solution, thus lowering its molarity:

$$\ce{\overset{0}{Mg} (s) + 2\overset{+1}{Ag}NO3 (aq) -> \overset{+2}{Mg}(NO3)2 (aq) + \overset{0}{Ag} (s)}$$

Final concentration $$c_2(\ce{AgNO3})$$ is pretty much defined by the remaining silver(I) nitrate when magnesium is depleted in the reaction:

$$c_2(\ce{AgNO3}) = c_1(\ce{AgNO3}) - \Delta c(\ce{AgNO3})\label{eq:1}\tag{1}$$

$$c_1(\ce{AgNO3})$$ is the known initial concentration; $$V$$ is the volume; $$\Delta c(\ce{AgNO3})$$ is the change in concentration:

$$\Delta c(\ce{AgNO3}) = \frac{\Delta n(\ce{AgNO3})}{V}\label{eq:2}\tag{2}$$

where the unknown amount of silver nitrate $$\Delta n(\ce{AgNO3})$$ can be found knowing stoichiometry of the reaction (assuming the reaction is complete and is irreversible), mass and molar mass of magnesium $$m(\ce{Mg})$$ and $$M(\ce{Mg})$$, respectively:

$$\Delta n(\ce{AgNO3}) = 2n(\ce{Mg}) = \frac{2m(\ce{Mg})}{M(\ce{Mg})}\label{eq:3}\tag{3}$$

At this point we can rewrite \eqref{eq:1} using \eqref{eq:2} and \eqref{eq:3} since all the variables are known:

$$c_2(\ce{AgNO3}) = c_1(\ce{AgNO3}) - \frac{2m(\ce{Mg})}{VM(\ce{Mg})} = \pu{0.625 M} - \frac{2\cdot\pu{3 g}}{\pu{0.500 L}\cdot\pu{24 g mol-1}} = \pu{0.125 M}$$