Question

What mass of Aluminum Hydroxide will be produced by the reaction of 78.33g of aluminum with excess water?

Answer 1

226.5g

Explanation:

First we need the balanced equation:

`2Al + 6H_2O -> 2Al(OH)_3 + 3H_2`

We then need to know how many moles is equivalent to 78.33g

`78.33g_"Al"((1 mol_"Al")/(26.98g_"Al")) = 2.903mol_"Al"`

Since 2 moles of `Al(OH)_3`

Is produced for every 2 moles of `Al`, they are equivalent, producing `2.903mol_"Al(OH)3"`.

Lastly, convert moles to grams.

`2.903mol_"Al(OH)3"((78.01g_"Al(OH)3")/(1 mol_"Al(OH)3"))` = `226.5g_"Al(OH)3"`