Asample 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?

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

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

answer; i believe that the correct answer is (a) potassium(k)