Question

write, complete, and balance the following then perform the stoichiometric calculations.

1. write the decomposition of aluminum hydroxide. if 255g of the metal oxide are produced, how grams of the original reactant did you start with?

Explanation:

Step1: Write unbalanced decomposition reaction

$\text{Al(OH)}_3
ightarrow \text{Al}_2\text{O}_3 + \text{H}_2\text{O}$

Step2: Balance the chemical equation

Balance Al: $2\text{Al(OH)}_3
ightarrow \text{Al}_2\text{O}_3 + \text{H}_2\text{O}$
Balance H and O: $2\text{Al(OH)}_3
ightarrow \text{Al}_2\text{O}_3 + 3\text{H}_2\text{O}$

Step3: Calculate molar masses

Molar mass of $\text{Al}_2\text{O}_3$: $2\times27 + 3\times16 = 102\ \text{g/mol}$
Molar mass of $\text{Al(OH)}_3$: $27 + 3\times(16+1) = 78\ \text{g/mol}$

Step4: Find moles of product

$\text{Moles of Al}_2\text{O}_3 = \frac{255\ \text{g}}{102\ \text{g/mol}} = 2.5\ \text{mol}$

Step5: Relate moles via stoichiometry

From balanced equation, 1 mol $\text{Al}_2\text{O}_3$ comes from 2 mol $\text{Al(OH)}_3$.
$\text{Moles of Al(OH)}_3 = 2.5\ \text{mol} \times 2 = 5\ \text{mol}$

Step6: Calculate mass of reactant

$\text{Mass of Al(OH)}_3 = 5\ \text{mol} \times 78\ \text{g/mol} = 390\ \text{g}$

Answer:

Balanced reaction: $2\text{Al(OH)}_3
ightarrow \text{Al}_2\text{O}_3 + 3\text{H}_2\text{O}$
Initial mass of aluminum hydroxide: 390 grams