# Am I doing this right? What mass of "HCl" reacts with "3.26 g" of "Mg"("OH")_2?

## $\text{3.26 g Mg"("OH")_2 xx ("1 mol Mg"("OH")_2)/("58.326 g Mg"("OH")_2) xx ("2 mol HCl")/("1 mol Mg"("OH")_2) xx "36.458 g HCl"/"1 mol HCl}$ $=$ $\text{4.075 g}$

Sep 25, 2016

Here's what I would do. Since $\text{3.26 g}$ of "Mg"("OH")_2 neutralizes a certain amount of $\text{HCl}$, we start with that much mass of magnesium hydroxide and determine what mass of $\text{HCl}$ it corresponds to.

The balanced equation is

$2 {\text{HCl"(aq) + "Mg"("OH")_2(s) -> 2"H"_2"O"(l) + "MgCl}}_{2} \left(a q\right)$

So we use the general pathway

$\text{g Mg"("OH")_2 stackrel(-: "g/mol Mg"("OH")_2)(->) "mols Mg"("OH")_2 stackrel(xx "mol HCl"/("mol Mg"("OH")_2))(->) "mols HCl" stackrel(xx "g/mol HCl")(->) "g HCl}$

to write out

3.26 cancel("g Mg"("OH")_2) xx cancel("1 mol Mg"("OH")_2)/(58.3188 cancel("g Mg"("OH")_2)) xx (2 cancel("mol HCl"))/cancel("1 mol Mg"("OH")_2) xx "36.4609 g HCl"/cancel"1 mol HCl"

$=$ $\textcolor{b l u e}{\text{4.08 g HCl}}$

It looks like your mathematical process is right, but you seem to have used slightly different molar masses.

I used $\text{35.453 g/mol Cl}$, $\text{1.0079 g/mol H}$, $\text{24.305 g/mol Mg}$ and $\text{15.999 g/mol O}$. Other than that, you have it.