Question

What volume of oxygen gas is produced when 14.3 g of mercury(II) oxide reacts completely according to the following reaction at 25 °C and 1 atm? Volume = ?

A) 3.56 L
B) 4.82 L
C) 2.91 L
D) 5.73 L

Answer 1

To determine the volume of oxygen gas produced when mercury(II) oxide reacts completely, we need to calculate the moles of mercury(II) oxide, use the balanced equation to find the moles of oxygen gas produced, and then use the ideal gas law to convert moles to volume. The volume of oxygen gas produced is approximately 5.73 L.

Step-by-step explanation:

To determine the volume of oxygen gas produced when mercury(II) oxide reacts completely, we need to first find the moles of mercury(II) oxide using its molar mass. Then, using the balanced chemical equation, we can calculate the moles of oxygen gas produced. Finally, using the ideal gas law, we can convert the moles of oxygen gas to volume.

Given: Mass of HgO = 14.3 g, Temperature = 25°C, Pressure = 1 atm

1. Calculate the moles of HgO: Moles of HgO = Mass of HgO / Molar mass of HgO
2. Use the balanced chemical equation to determine the moles of O₂ produced: Moles of O₂ = Moles of HgO × Moles of O₂ (from the equation)
3. Apply the ideal gas law to calculate the volume of O₂: Volume of O₂ = (Moles of O₂ × Gas constant × Temperature) / Pressure

The calculated volume of oxygen gas produced is approximately 5.73 L, which corresponds to option D) 5.73 L.