Oxygen can be produced by the thermal
decomposition of mercuric oxide:
2 HgO(s) → 2 Hg(l) + O2(g)
How many liters of O2 are produced at
50.0°C and 0.947 atm by decomposition of
57.0 g of HgO?
To determine the number of liters of oxygen (O2) produced by the thermal decomposition of mercuric oxide (HgO), we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas (in atm),
V is the volume of the gas (in liters),
n is the number of moles,
R is the ideal gas constant (0.0821 L*atm/(mol*K)),
T is the temperature of the gas (in Kelvin).
To solve this problem, we need to follow these steps:
Step 1: Convert the given temperature (50.0°C) to Kelvin.
The temperature in Kelvin can be obtained by adding 273.15 to the temperature in degrees Celsius.
T(K) = T(°C) + 273.15
T(K) = 50.0 + 273.15 = 323.15 K
Step 2: Calculate the number of moles of HgO.
To find the number of moles (n), we will use the molar mass of HgO, which is 216.59 g/mol.
n = mass / molar mass
n = 57.0 g / 216.59 g/mol
n = 0.263 moles of HgO
Step 3: Use the stoichiometry of the reaction to determine the number of moles of O2 produced.
From the balanced equation, we can see that 2 moles of HgO decompose to produce 1 mole of O2.
So, for 0.263 moles of HgO, the number of moles of O2 produced would be:
(0.263 moles HgO) * (1 mole O2 / 2 moles HgO) = 0.132 moles O2
Step 4: Apply the ideal gas law equation to calculate the volume of O2 produced.
PV = nRT
Rearranging the formula, we get:
V = (nRT) / P
V = (0.132 moles) * (0.0821 L*atm/(mol*K)) * (323.15 K) / 0.947 atm
V ≈ 3.55 L
Therefore, approximately 3.55 liters of O2 are produced at 50.0°C and 0.947 atm by the decomposition of 57.0 g of HgO.
To find the number of liters of O2 produced, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Gas constant (0.0821 L.atm/mol.K)
T = Temperature (in Kelvin)
First, we need to calculate the number of moles of HgO.
We can use the molar mass of HgO to convert grams to moles:
molar mass of HgO = 200.59 g/mol
moles of HgO = mass of HgO / molar mass of HgO
moles of HgO = 57.0 g / 200.59 g/mol
moles of HgO = 0.284 mol
From the balanced equation, we can see that for every 2 moles of HgO, 1 mole of O2 is produced.
So, the number of moles of O2 produced is half the number of moles of HgO:
moles of O2 = 0.5 * moles of HgO
moles of O2 = 0.5 * 0.284 mol
moles of O2 = 0.142 mol
Next, we need to convert the temperature from Celsius to Kelvin:
T (K) = T (°C) + 273.15
T (K) = 50.0°C + 273.15
T (K) = 323.15 K
Now, we can use the ideal gas law equation to find the volume of O2:
V = (nRT) / P
V = (0.142 mol * 0.0821 L.atm/mol.K * 323.15 K) / 0.947 atm
V ≈ 3.72 L
Therefore, approximately 3.72 liters of O2 are produced at 50.0°C and 0.947 atm by the decomposition of 57.0 g of HgO.