Question

HgO is decomposed into Hg and O2. What volume of O2 can be produced?

A) Stoichiometric Volume
B) Oxygen Yield
C) Decomposition Ratio
D) Gas Generation

Answer 1

To calculate the volume of oxygen gas produced from the decomposition of HgO, find the moles of HgO decomposed, use the stoichiometry of the reaction to determine the moles of O2 produced, and then apply the Ideal Gas Law to find the oxygen volume at specific temperature and pressure, which in this case is approximately 0.308 liters.

Step-by-step explanation:

To calculate the volume of O2 produced from the decomposition of mercury(II) oxide (HgO), follow these steps:

1. Determine the moles of HgO that decompose by using its molar mass.
2. Using the chemical equation, establish the stoichiometric relationship between HgO and O2 to find the moles of oxygen produced.
3. Use the Ideal Gas Law (PV=nRT) to calculate the volume of O2 at the given temperature and pressure.

For the decomposition of 5.36 g of HgO:

1. Molar mass of HgO is approximately 216.59 g/mol. Therefore, moles of HgO = 5.36 g / 216.59 g/mol ≈ 0.0247 mol.
2. The balanced chemical equation for the decomposition of HgO is 2HgO → 2Hg + O2, which shows a 2:1 mole ratio. Therefore, the moles of O2 produced = 0.0247 mol HgO * (1 mol O2/2 mol HgO) = 0.01235 mol.
3. Assuming ideal gas behavior, at 23 °C (which is 296.15 K) and 0.975 atm, the volume V can be calculated using the formula PV=nRT, where R is the gas constant (0.0821 L·atm/mol·K). V = (nRT)/P = (0.01235 mol * 0.0821 L·atm/mol·K * 296.15 K) / 0.975 atm ≈ 0.308 L of O2.