Question

1. if you start with 65 grams of $\text{mg(clo}_3\text{)}_2$, how many moles of $\text{o}_2$ will be produced?

$\text{mg(clo}_3\text{)}_2
ightarrow \text{mgcl}_2 + 3\text{o}_2$

Explanation:

Step1: Calculate molar mass of $\text{Mg(ClO}_3\text{)}_2$

Molar mass = $24.31 + 2\times(35.45 + 3\times16.00) = 191.21\ \text{g/mol}$

Step2: Convert mass to moles of $\text{Mg(ClO}_3\text{)}_2$

$\text{Moles of Mg(ClO}_3\text{)}_2 = \frac{65\ \text{g}}{191.21\ \text{g/mol}} \approx 0.340\ \text{mol}$

Step3: Use mole ratio to find $\text{O}_2$ moles

From balanced equation, 1 mol $\text{Mg(ClO}_3\text{)}_2$ produces 3 mol $\text{O}_2$.
$\text{Moles of O}_2 = 0.340\ \text{mol} \times 3 = 1.02\ \text{mol}$

Answer:

Approximately 1.0 moles (or 1.02 moles) of $\text{O}_2$ will be produced.