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The balanced chemical reaction is expressed as:

3HCl + Al(OH)3 = AlCl3 + 3H2O

We are given the amount of aluminum hydroxide to be used up in the reaction. This value will be the starting point for the calculations. We calculate as follows:

1.55 g Al(OH)3 ( 1 mol Al(OH)3 / 78 g Al(OH)3 )( 3 mol HCl / 1 mol Al(OH)3 ) = 0.06 mol HCl

Volume = 0.06 mol HCl / 0.120 mol HCl/ L solution = 0.50 L solution

Answer: The volume of HCl is 55.83 mL

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass of aluminium hydroxide = 1.55 g

Molar mass of aluminium hydroxide = 78 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of aluminium hydroxide}=\frac{1.55g}{78g/mol}=0.020mol[/tex]

The chemical reaction for the formation of chromium oxide follows the equation:

[tex]Al(OH)_3+3HCl\rightarrow AlCl_3+6H_2O[/tex]

By Stoichiometry of the reaction:

1 mole of aluminium hydroxide reacts with 3 moles of HCl

So, 0.020 moles of aluminium hydroxide will react with = [tex]\frac{1}{3}\times 0.020=0.0067mol[/tex] of HCl

To calculate the volume of HCl, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]

We are given:

Moles of HCl = 0.0067 moles

Molarity of solution = 0.120 M

Putting values in above equation, we get:

[tex]0.120=\frac{0.0024\times 1000}{\text{Volume of HCl}}\\\\\text{Volume of HCl}=\frac{0.0024\times 1000}{0.120}=55.83mL[/tex]

Hence, the volume of HCl is 55.83 mL

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