Question

1. find the volume of 0.110 mol l-1 hydrochloric acid necessary to react completely with 1.52 g al(oh)3.

a) 0.658 l
b) 1.52 l
c) 1.88 l
d) 0.530 l
e) 1.01 l

Explanation:

Step1: Calculate moles of Al(OH)₃

The molar mass of Al(OH)₃ is \(M = 27+(16 + 1)\times3=78\ g/mol\). The number of moles of Al(OH)₃, \(n_{Al(OH)_3}=\frac{m}{M}=\frac{1.52\ g}{78\ g/mol}\approx0.0195\ mol\).

Step2: Determine mole - ratio from chemical equation

The balanced chemical equation for the reaction between HCl and Al(OH)₃ is \(3HCl + Al(OH)_3=AlCl_3 + 3H_2O\). The mole - ratio of HCl to Al(OH)₃ is \(n_{HCl}:n_{Al(OH)_3}=3:1\). So the number of moles of HCl required, \(n_{HCl}=3\times n_{Al(OH)_3}=3\times0.0195\ mol = 0.0585\ mol\).

Step3: Calculate volume of HCl

We know that \(n = cV\) (where \(n\) is the number of moles, \(c\) is the concentration, and \(V\) is the volume). Given \(c = 0.110\ mol/L\) and \(n = 0.0585\ mol\), we can solve for \(V\). \(V=\frac{n}{c}=\frac{0.0585\ mol}{0.110\ mol/L}\approx0.530\ L\).

Answer:

D. 0.530 L