Question

When chromium (III) chloride is combined with sodium hydroxide, it forms chromium (III) hydroxide (a colourful green precipitate). The reaction is represented as follows:
$CrC{l}_{3\left(aq\right)}+NaO{H}_{\left(aq\right)}\to Cr(OH{)}_{3\left(s\right)}+NaC{l}_{\left(aq\right)}$
When the equation is balanced, how many moles of chromium (III) hydroxide will be formed?

• A. $1mole$
• B. $2moles$
• C. $3moles$
• D. $4moles$

Answer 1

The correct option is A $1mole$

$CrC{l}_3+NaOH\to Cr(OH{)}_3+NaCl(aq)$

Products has $3moles$ of oxygen, so reactant $NaOH$ should have coefficient $3.$

$CrC{l}_3+3NaOH\to Cr(OH{)}_3+NaCl(aq)$

As reactants has $3moles$ of chlorine, product $NaCl$ should have coefficient $3.$

$CrC{l}_3+3NaOH\to Cr(OH{)}_3+3NaCl(aq)$

So, $1mole$ of $CrC{l}_3$ should produce $1mole$ of $Cr(OH{)}_3.$