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

"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"

= "4.075 g"

1 Answer
Sep 25, 2016

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

The balanced equation is

2"HCl"(aq) + "Mg"("OH")_2(s) -> 2"H"_2"O"(l) + "MgCl"_2(aq)

So we use the general pathway

"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"

= color(blue)("4.08 g HCl")

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

I used "35.453 g/mol Cl", "1.0079 g/mol H", "24.305 g/mol Mg" and "15.999 g/mol O". Other than that, you have it.