# Question #1eec5

Jan 1, 2017

You need to use 76.9 mL of $\text{HCl}$.

#### Explanation:

Step 1. Start with the equation for the reaction

${M}_{r} : \textcolor{w h i t e}{m m m m m m m m l} 78.00$
$\textcolor{w h i t e}{m m m m m m} \text{3HCl" + "Al(OH)"_3 → "AlCl"_3 + "3H"_2"O}$

Step 2. Calculate the moles of ${\text{Al(OH)}}_{3}$

${\text{Moles of Al(OH)"_3 = 1.00 color(red)(cancel(color(black)("g Al(OH)"_3))) × ("1 mol Al(OH)"_3)/(78.00 color(red)(cancel(color(black)("g Al(OH)"_3)))) = "0.012 82 mol Al(OH)}}_{3}$

Step 3. Calculate the moles of HCl needed

$\text{Moles of HCl" = "0.012 82" color(red)(cancel(color(black)("mol Al(OH)"_3))) × "3 mol HCl"/(1 color(red)(cancel(color(black)("mol Al(OH)"_3)))) = "0.038 46 mol HCl}$

Step 4. Calculate the volume of HCl

$\text{Volume of HCl" = "0.038 46" color(red)(cancel(color(black)("mol HCl"))) × "1 L HCl"/(0.500 color(red)(cancel(color(black)("mol HCl")))) = "0.076 92 L HCl" = "76.9 mL HCl}$