Final answer:
To calculate the volume of oxygen gas produced from 5.0 moles of mercury(II) oxide, we find 2.5 moles of oxygen produced and convert this to grams, and then to liters using the given density. The result is 55.6 liters of oxygen gas.
Step-by-step explanation:
The question asks us to calculate the volume of oxygen gas produced when 5.0 moles of mercury(II) oxide are heated. To find this, we use the balanced chemical equation for the decomposition of mercury(II) oxide, which is:
2 HgO (s) → 2 Hg (l) + O₂ (g).
From the equation, we see that 2 moles of HgO produce 1 mole of O₂ gas. Therefore, 5.0 moles of HgO will produce 2.5 moles of O₂ (since 5.0 divided by 2 equals 2.5).
To find the volume, we use the molar volume of a gas under standard conditions (22.4 L/mol at STP), but since we are given the density instead, we can directly calculate the volume at the given conditions of 1.439 g/L.
First, we find the mass of 2.5 moles of O₂ which is 2.5 moles × 32.00 g/mole = 80.00 g of O₂.
Then we divide the mass by the density to find the volume: 80.00 g / 1.439 g/L = 55.6 L.
Thus, 55.6 liters of oxygen gas can be produced by heating 5.0 moles of mercury(II) oxide.