Answer :
Approximately 18.0 liters of H₂ gas are needed.
Here's how to find the volume of H₂ gas needed:
**1. Balanced Chemical Equation:**
First, write the balanced chemical equation for the reaction:
CuO(s) + H₂(g) → Cu(s) + H₂O(g)
**2. Moles of CuO:**
Calculate the moles of CuO:
- Molar mass of CuO: 79.55 g/mol
- Moles of CuO: 35.5 g / 79.55 g/mol ≈ 0.446 mol
**3. Stoichiometry and Moles of H₂:**
According to the balanced equation, 1 mole of CuO reacts with 1 mole of H₂. Therefore, you need 0.446 mol of H₂.
**4. Ideal Gas Law and Volume:**
Use the ideal gas law to relate moles, pressure, volume, and temperature:
PV = nRT
where:
- P: pressure (765 torr) = 765 * (1 atm / 760 torr) ≈ 1 atm
- V: volume (unknown)
- n: moles of H₂ (0.446 mol)
- R: gas constant (0.08206 L atm/mol K)
- T: temperature (225°C + 273.15 K) = 498.15 K
**5. Solve for Volume:**
Rearrange the equation to solve for V:
V = nRT / P
V ≈ (0.446 mol * 0.08206 L atm/mol K * 498.15 K) / 1 atm
V ≈ 18.0 L
Therefore, approximately 18.0 liters of H₂ gas are needed.
The probable question may be:
What volume of H2 gas at 765 torr and 225°C is needed to reduce 35.5 g of copper(II)oxide (79.55 g/mol) to form pure copper and water? Round your answer to one decimal place and do not include units.
