Question

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?

Answer 1

`80` `"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 `"Al(OH)"_3` that react, knowing that `1` `"mol AlCl"_3` forms:

`1cancel("mol AlCl"_3)((1color(white)(l)"mol Al(OH)"_3)/(1cancel("mol AlCl"_3))) = 1` `"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` `"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..