# The balanced equation shows the reaction between aluminium hydroxide and hydrochloric acid to form aluminium chloride and water: Al(OH)3 + 3HCl = AlCl3 + 3H2O. 1 mole of aluminium chloride was formed, what mass of aluminium hydroxide was formed?

Jul 9, 2017

$80$ ${\text{g Al(OH)}}_{3}$ (one sig fig)

#### Explanation:

I'll assume you mean what mass of aluminum hydroxide reacted.

We can use the coefficients of the equation to determine the number of moles of ${\text{Al(OH)}}_{3}$ that react, knowing that $1$ ${\text{mol AlCl}}_{3}$ forms:

1cancel("mol AlCl"_3)((1color(white)(l)"mol Al(OH)"_3)/(1cancel("mol AlCl"_3))) = 1 ${\text{mol Al(OH)}}_{3}$

which makes sense given their molar ratio is $1 : 1$.

Now, we can use the molar mass of aluminum hydroxide ($78.00$ $\text{g/mol}$) to calculate the number of grams that react:

1cancel("mol Al(OH)"_3)((78.00color(white)(l)"g Al(OH)"_3)/(1cancel("mol Al(OH)"_3))) = color(red)(80 color(red)("g Al(OH)"_3

which I suppose will round to $1$ significant figure, the amount given in the problem..