Question d7c42

Feb 3, 2015

The answer is $\text{0.596 L}$ or $\text{596 mL}$.

Explanation:

${\text{Al"("OH")_ (3(s)) + 3"HCl"_ ((aq)) -> "AlCl"_ (3(aq)) + 3"H"_ 2"O}}_{\left(l\right)}$

Look at the mole ratio you have between aluminium hydroxide and hydrochloric acid--$1$ mole of the former will react with $3$ moles of the latter, i.e. you have $1 : 3$ mole ratio between the two compounds.

This means that you need three times as many moles of hydrochloric acid as you have of aluminium hydroxide. SInce you know how many moles of aluminium oxide you have--because you were given the mass of the sample, you can easily determine how many moles of hydrochloric would be needed.

Use the molar mass of aluminium oxide to get

1.55 cancel("g") * ("1 mole Al"("OH")_3)/(78.0 cancel("g")) = "0.01987 moles"

As a result, you need

0.01987 cancel("moles Al"("OH")_3) * ("3 moles HCl")/(1cancel("moles Al"("OH")_3)) = "0.05961 moles HCl"

Now just use the definition of molarity--moles of solute divided by volume of solution. You must supply these $\text{0.05961 moles}$ of hydrochloric acid to the reaction by means of a solution in which hydrochloric acid is the solute.

So,

c = n/V => V = n/c = ("0.05961 moles")/("0.100 mol/L") = "0.596 L" -> 3 sig figs

So $\text{0.596 L}$ of $\text{0.100 M}$ hydrochloric acid solution will provide the number of moles needed to completely react with that mass of "Al"("OH")_3#.