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) 0.531 l
b) 0.177 l
c) 5.65 l
d) 0.0590 l
e) 0.314 l

Explanation:

Step1: Calculate moles of Al(OH)₃

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

Step2: Determine mole - ratio from balanced 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 3:1. So, $n_{HCl}=3\times n_{Al(OH)_3}=3\times0.0195\ mol = 0.0585\ mol$

Step3: Calculate volume of HCl

We know that $M=\frac{n}{V}$, so $V=\frac{n}{M}$. Given $M = 0.110\ mol/L$ and $n = 0.0585\ mol$. Then $V=\frac{0.0585\ mol}{0.110\ mol/L}\approx0.531\ L$

Answer:

A. 0.531 L