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.