Question

1. mg(clo₃)₂ → mgcl₂ + 3o₂

calculate the mass of oxygen gas formed from the decomposition of 2.39 g of magnesium chlorate.

Explanation:

Step1: Balance the chemical equation

The unbalanced equation is \( \text{Mg(ClO}_3\text{)}_2
ightarrow \text{MgCl}_2 + \text{O}_2 \).
To balance O, we see that in \( \text{Mg(ClO}_3\text{)}_2 \), there are 6 O atoms. So we put a coefficient of 3 in front of \( \text{O}_2 \), and the balanced equation is:
\( \text{Mg(ClO}_3\text{)}_2
ightarrow \text{MgCl}_2 + 3\text{O}_2 \)

Step2: Calculate moles of \( \text{Mg(ClO}_3\text{)}_2 \)

Molar mass of \( \text{Mg(ClO}_3\text{)}_2 \):
Mg: 24.31 g/mol, Cl: 35.45 g/mol, O: 16.00 g/mol
Molar mass = \( 24.31 + 2\times(35.45 + 3\times16.00) \)
= \( 24.31 + 2\times(35.45 + 48.00) \)
= \( 24.31 + 2\times83.45 \)
= \( 24.31 + 166.9 \)
= \( 191.21 \) g/mol

Moles of \( \text{Mg(ClO}_3\text{)}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{2.39\ \text{g}}{191.21\ \text{g/mol}} \approx 0.0125\ \text{mol} \)

Step3: Use mole ratio to find moles of \( \text{O}_2 \)

From the balanced equation, 1 mol of \( \text{Mg(ClO}_3\text{)}_2 \) produces 3 mol of \( \text{O}_2 \).
So moles of \( \text{O}_2 = 0.0125\ \text{mol} \times 3 = 0.0375\ \text{mol} \)

Step4: Calculate mass of \( \text{O}_2 \)

Molar mass of \( \text{O}_2 = 32.00\ \text{g/mol} \)
Mass of \( \text{O}_2 = \text{moles} \times \text{molar mass} = 0.0375\ \text{mol} \times 32.00\ \text{g/mol} = 1.2\ \text{g} \) (rounded to two decimal places)

Answer:

The mass of oxygen gas formed is approximately \( 1.20\ \text{g} \) (or 1.2 g depending on significant figures).