Answer :
Approximately 0.086 liters of oxygen gas would be produced when 51.7 g of mercury(II) oxide reacts completely at 25 °C and 1 atm.
To determine the volume of oxygen gas produced, we need to use the stoichiometry of the balanced chemical equation and the ideal gas law. First, let's write the balanced equation for the reaction:
2 HgO (s) → 2 Hg (l) + O2 (g)
From the equation, we can see that 2 moles of HgO produce 1 mole of O2.
Given:
Mass of HgO = 51.7 g
Molar mass of HgO = 200.59 g/mol
We can calculate the number of moles of HgO:
moles of HgO = mass / molar mass
moles of HgO = 51.7 g / 200.59 g/mol
Next, we use the stoichiometric ratio to determine the number of moles of O2 produced:
moles of O2 = (moles of HgO) / 2
Now, we can use the ideal gas law equation to find the volume of gas:
PV = nRT
Rearranging the equation to solve for volume (V):
V = (nRT) / P
Where:
P = pressure = 1 atm
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature = 25 °C = 25 + 273.15 K = 298.15 K
Substituting the values, we get:
V = (moles of O2 * R * T) / P
V = ((moles of HgO) / 2) * 0.0821 L·atm/(mol·K) * 298.15 K / 1 atm
Now, we can substitute the value for moles of HgO:
V = (51.7 g / 200.59 g/mol / 2) * 0.0821 L·atm/(mol·K) * 298.15 K / 1 atm
Calculating the volume of oxygen gas:
V ≈ 0.086 L
Therefore, approximately 0.086 liters of oxygen gas would be produced when 51.7 g of mercury(II) oxide reacts completely at 25 °C and 1 atm.
To know more about ideal gas, visit:
https://brainly.com/question/30236490
#SPJ11
