Question #1eec5

1 Answer
Jan 1, 2017

Answer:

You need to use 76.9 mL of #"HCl"#.

Explanation:

Step 1. Start with the equation for the reaction

#M_r:color(white)(mmmmmmmml)78.00#
#color(white)(mmmmmm)"3HCl" + "Al(OH)"_3 → "AlCl"_3 + "3H"_2"O"#

Step 2. Calculate the moles of #"Al(OH)"_3#

#"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

#"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

#"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"#