Question

What mass of HCL, in grams, is required to react with 0.610 g of al(oh)3 ?

Answer 1

Answer: 0.8541 grams of HCl will be required.

Explanation: Moles can be calculated by using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $Al(OH)_3$ = 0.610 g

Molar mass of $Al(OH)_3$ = 78 g/mol

$\text{Number of moles}=\frac{0.610g}{78g/mol}$

Number of moles of $Al(OH)_3$ = 0.0078 moles

The reaction between $Al(OH)_3$ and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.

Chemical equation for the above reaction follows:

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

By Stoichiometry,

1 mole of  $Al(OH)_3$ reacts with 3 moles of HCl

So, 0.0078 moles of $Al(OH)_3$ will react with $\frac{3}{1}\times 0.0078$ = 0.0234 moles

Mass of HCl is calculated by using the mole formula, we get

Molar mass of HCl = 36.5 g/mol

Putting values in the equation, we get

$0.0234moles=\frac{\text{Given mass}}{36.5g/mol}$

Mass of HCl required will be = 0.8541 grams