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What volume of oxygen at 439.0K and a pressure of 1.00 atm is produced by the decomposition of 134.0g of BaO₂? Use a molar mass of 169.33 g/mol for BaO₂.

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Final answer:

The volume of oxygen produced by the decomposition of BaO₂ can be calculated by first determining the moles of BaO₂ decomposed, then using stoichiometry to find the moles of O₂, and finally applying the ideal gas law with the given temperature and pressure.

Step-by-step explanation:

Calculating the Volume of Oxygen Produced by Decomposition

To determine the volume of oxygen gas produced by the decomposition of barium peroxide (BaO₂), we first need to find the number of moles of BaO₂ that decomposed. The molar mass of BaO₂ is given as 169.33 g/mol. Using the mass given in the problem, we can calculate the moles of BaO₂:

moles of BaO₂ = mass / molar mass = 134.0 g / 169.33 g/mol

Next, the balanced equation for the decomposition of BaO₂ to BaO and O₂ is:

2 BaO₂(s) → 2 BaO(s) + O₂(g)

From the stoichiometry of the equation, we see that 2 moles of BaO₂ produce 1 mole of O₂, thus:

moles of O₂ = moles of BaO₂ / 2

Now, we use the ideal gas law to calculate the volume of O₂ produced:

PV = nRT

Where:

  • P is the pressure in atmospheres (atm)
  • V is the volume in liters (L)
  • n is the number of moles of gas
  • R is the ideal gas constant (0.0821 L·atm/K·mol)
  • T is the absolute temperature in kelvins (K)

V = (nRT) / P

After calculating the moles of O₂, insert the values into the ideal gas law equation along with R = 0.0821 L·atm/K·mol, T = 439.0 K, and P = 1.00 atm to find the volume of oxygen.

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