What is the number of moles of hydrochloric acid that will be neutralized if there are 750 mg of aluminum hydroxide are consumed? What is the mass of water that will be produced?

Equation: #"Al"("OH")_3(s) + 3"HCl"(aq) -> "AlCl"_3(aq) + 3"H"_2"O"(l)#

This is what I think I know so far: I have to convert the 750 mg to g, which is .75 g. I believe that I have to find the number of moles in .75 g of Al(OH)3, and then multiply that number by 3, since the ratio of Al(OH)3 to HCl is 1:3.

1 Answer
Dec 21, 2017


You are on the right track.


Indeed, your approach here will be to use the molar mass of aluminium hydroxide to convert the mass of the sample to moles and the mole ratio that exists between the two reactants to find the number of moles of hydrochloric acid consumed.

#"Al"("OH")_ (3(s)) + 3"HCl"_ ((aq)) -> "AlCl"_ (3(aq)) + 3"H"_ 2"O"_ ((l))#

The balanced chemical equation tells you that the reaction consumes #3# moles of hydrochloric acid and produces #3# moles of water for every #1# mole of aluminium hydroxide that takes part in the reaction.

So you can say that you have

#0.75 color(red)(cancel(color(black)("g"))) * overbrace(("1 mole Al"("OH")_3)/(78 color(red)(cancel(color(black)("g")))))^(color(blue)("the molar mass of Al"("OH")_3)) * overbrace("3 moles HCl"/(1color(red)(cancel(color(black)("mole Al"("OH")_3)))))^(color(blue)("the 1:3 mole ratio")) = color(darkgreen)(ul(color(black)("0.029 moles HCl")))#

Since you know that the reaction produces the same number of moles of water as the number of moles of hydrochloric acid it consumes--the two chemical species have a #3:3# mole ratio in the balanced chemical equation--you can use the molar mass of water to say that the reaction will produce

#0.029 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("0.52 g")))#

The answers are rounded to two sig figs, the number of sig figs you have for the mass of aluminium hydroxide.