How many liters of hydrogen,measured at 20°C and 750 torr,are required to react with 55.2 grams of aluminum oxide in the reaction Al₂O₃ + 3 H₂ → 2 Al + 3 H₂O?

Question

Quizplus · Accepted Answer

To solve this, first, we need to balance the chemical equation, which is already given as: $$Al_2O_3 + 3H_2 \rightarrow 2Al + 3H_2O$$. Next, we calculate the moles of $Al_2O_3$ using its molar mass (101.96 g/mol for $Al_2O_3$): $$\frac{55.2 \text{ g}}{101.96 \text{ g/mol}} = 0.541 \text{ mol}$$. According to the balanced equation, 1 mole of $Al_2O_3$ reacts with 3 moles of $H_2$, so $0.541 \text{ mol}$ of $Al_2O_3$ will require $0.541 \times 3 = 1.623 \text{ mol}$ of $H_2$. Using the ideal gas law $PV = nRT$ and converting conditions to SI units (750 torr to atm and 20°C to 293K), we find the volume of $H_2$: $$V = \frac{nRT}{P} = \frac{1.623 \text{ mol} \times 0.0821 \text{ L·atm·K}^{-1}\text{mol}^{-1} \times 293 \text{ K}}{750 \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}}} \approx 36.6 \text{ L}$$.

How Many Liters of Hydrogen,measured at 20°C and 750 Torr,are