Question

A sample of mercury(II) oxide is placed in a 5.00 L evacuated container and heated until it decomposes entirely to mercury metal and oxygen gas. The container is then cooled to 25°C. One now finds that the gas pressure inside the container is 1.73 atm. What mass of mercury(II) oxide was originally placed into the container?

Answer 1

There was originally placed 153 grams of mercury(II) oxide in the container

Step-by-step explanation:

Step 1: The balanced equation

2HgO → 2Hg + O2

Step 2: Data given

Volume of the container = 5.00L

Temperature in the container = 25°C = 298.15 Kelvin

The gas pressure inside = 1.73 atm

Step 3: Calculate number of moles of O2

We use the ideal gas law for this

p*V = n * R * T

with p = the pressure of the gas = 1.73 atm

with V = the volume of the gas = 5.00 L

with n = the number of moles of the gas = TO BE DETERMINED

with R = the gas constant = 0.0821 L*atm*K^−1*mol^−1

T = The temperature in the container = 25°C = 298.15 Kelvin

p*V = n * R *T

n = P*V / R*T

n = 1.73 atm * 5.00 L / (0.0821 L*atm*K^−1*mol^−1 * 298.15 K)

n = 0.353 moles O2

Step 4: Calculate moles of HgO

In the equation 2HgO → 2Hg + O2 we can see that for 1 mole O2 produced, we need to consume 2 moles of HgO

if there is produced 0.353 moles of O2, we need to consume 2*0.353 = 0.706 moles of HgO

Step 5: Calculate mass of HgO

Mass of HgO = number of moles of HgO * Molar mass of HgO

Mass of HgO = 0.706 moles * 216.591 g/mole = 152.9 grams ≈ 153 grams of HgO

There was originally placed 153 grams of mercury(II) oxide in the container