Answer:
There was originally placed 153 grams of mercury(II) oxide in the container
Explanation:
<u>Step 1:</u> The balanced equation
2HgO → 2Hg + O2
<u>Step 2:</u> Data given
Volume of the container = 5.00L
Temperature in the container = 25°C = 298.15 Kelvin
The gas pressure inside = 1.73 atm
<u>Step 3: </u>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
<u>Step 4</u>: 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
<u>Step 5:</u> 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
