Question

What is the maximum mass of Al(OH)${}_3$ that can be prepared by reaction of 13.4 grams of AlCl${}_3$ with 10 grams of NaOH according to the following equation?
${\text{AlCl}}_{\text{3}}\text{+ 3NaOH}\to \text{Al}{\left(\text{OH}\right)}_{\text{3}}\text{+ 3NaCl}$

Answer 1

Moles of $AlC{l}_3=\frac{13.4}{133.5}=0.1$ $moles$

Moles of $NaOH=\frac{10}{40}=0.25$ $moles$

$AlC{l}_3+3NaOH⟶Al(OH{)}_3+3NaCl$

$0.1$ $0.25$

$0.0167$ $0.0833$

Since, $NaOH$ is limiting $0.0833$ $moles$ of $Al(OH{)}_3$ are formed.

Mass $=0.0833\times 78=6.5$ $gm$