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Is 225°C needed to reduce 35.5 g of copper (II) oxide?

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.

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