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?

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Answer 1 · 2017-07-09T22:49:38

$80$ $"g Al(OH)"_3$ (one sig fig)

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..