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2 H2O2(l) 2 H2O(g) + O2(g) Decompose 0.11 g of H2O2 in a flask with a volume of 2500.0 mL. What is the pressure of O2 at 25 oC? Of H2O?

Question

2 H2O2(l) 2 H2O(g) + O2(g)

Decompose 0.11 g of H2O2 in a flask with a volume of 2500.0 mL. What is the pressure of O2 at 25 oC? Of H2O?

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Solution

To solve this problem, we need to follow these steps:

  1. Convert the mass of H2O2 to moles. The molar mass of H2O2 is approximately 34.01 g/mol. So, 0.11 g of H2O2 is 0.11 g / 34.01 g/mol = 0.00323 mol.

  2. According to the balanced chemical equation, 2 moles of H2O2 decompose to produce 1 mole of O2. Therefore, 0.00323 mol of H2O2 will produce 0.00323 mol / 2 = 0.001615 mol of O2.

  3. Use the ideal gas law to find the pressure of O2. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

    Convert the temperature from Celsius to Kelvin: 25°C + 273.15 = 298.15 K.

    Convert the volume from mL to L: 2500.0 mL = 2.5 L.

    The ideal gas constant R is 0.0821 L·atm/(mol·K) when the pressure is in atmospheres.

    So, the pressure P of O2 is P = nRT / V = (0.001615 mol * 0.0821 L·atm/(mol·K) * 298.15 K) / 2.5 L = 0.0158 atm.

  4. According to the balanced chemical equation, 2 moles of H2O2 decompose to produce 2 moles of H2O. Therefore, 0.00323 mol of H2O2 will produce 0.00323 mol of H2O.

  5. Use the ideal gas law again to find the pressure of H2O. The number of moles n is 0.00323 mol. So, the pressure P of H2O is P = nRT / V = (0.00323 mol * 0.0821 L·atm/(mol·K) * 298.15 K) / 2.5 L = 0.0316 atm.

So, the pressure of O2 is 0.0158 atm and the pressure of H2O is 0.0316 atm.

This problem has been solved

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