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what quantity in moles of mno₃ are produced when 4.30 kg of oxygen gas …

Question

what quantity in moles of mno₃ are produced when 4.30 kg of oxygen gas completely reacts according to the balanced chemical reaction: 2 mn(s) + 3 o₂(g) → 2 mno₃(s)

Explanation:

Step1: Convert kg to g

$4.30\ \text{kg} \times 1000\ \text{g/kg} = 4300\ \text{g}$

Step2: Find moles of $\text{O}_2$

Molar mass of $\text{O}_2 = 32.00\ \text{g/mol}$
$\text{Moles of O}_2 = \frac{4300\ \text{g}}{32.00\ \text{g/mol}} = 134.375\ \text{mol}$

Step3: Mole ratio $\text{MnO}_3:\text{O}_2$

From reaction: $\frac{2\ \text{mol MnO}_3}{3\ \text{mol O}_2}$

Step4: Calculate moles of $\text{MnO}_3$

$134.375\ \text{mol O}_2 \times \frac{2\ \text{mol MnO}_3}{3\ \text{mol O}_2} \approx 89.6\ \text{mol}$

Answer:

$89.6$ moles