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

Apr 25, 2018

226.5g

#### Explanation:

First we need the balanced equation:

$2 A l + 6 {H}_{2} O \to 2 A l {\left(O H\right)}_{3} + 3 {H}_{2}$

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

$78.33 {g}_{\text{Al"((1 mol_"Al")/(26.98g_"Al")) = 2.903mol_"Al}}$

Since 2 moles of $A l {\left(O H\right)}_{3}$

Is produced for every 2 moles of $A l$, they are equivalent, producing $2.903 m o {l}_{\text{Al(OH)3}}$.

Lastly, convert moles to grams.

2.903mol_"Al(OH)3"((78.01g_"Al(OH)3")/(1 mol_"Al(OH)3")) = $226.5 {g}_{\text{Al(OH)3}}$