What volume of carbon dioxide would be produced at 350˚C and
1.00 atm from the complete thermal decomposition of 1.00 kg of
magnesium carbonate, MgCO3? What is the density of carbon dioxide
at this temperature?
MgCO3 ==> MgO + CO2
1 kg MgCO3 = 1000 g.
mols MgCO3 = grams/molar mass = ?
For every mol MgCO3 you get 1 mol CO2; therefore ?mols MgCO3 = the same number of mols of CO2 produced. Now convert that to the conditions you want using PV = nRT.
P = 1 atm
V = solve for this in L.
n = from above
R = 0.08206 L*atm/mol*K
T = 350 C but convert to kelvin. K = 273 + C
Post your work if you get stuck.
To find the volume of carbon dioxide produced from the thermal decomposition of magnesium carbonate, we can use the ideal gas law and the stoichiometry of the reaction.
The balanced equation for the thermal decomposition of magnesium carbonate is:
MgCO3(s) -> MgO(s) + CO2(g)
According to the equation, 1 mol of magnesium carbonate (MgCO3) produces 1 mol of carbon dioxide (CO2).
Step 1: Calculate the number of moles of magnesium carbonate.
1 kg of magnesium carbonate is equal to its molar mass.
The molar mass of MgCO3 = 24.31 g/mol (Mg) + 12.01 g/mol (C) + 3 * 16.00 g/mol (O)
= 84.31 g/mol
1 kg = 1000 g
Number of moles of MgCO3 = mass / molar mass
= 1000 g / 84.31 g/mol
≈ 11.87 mol
Step 2: Calculate the moles of CO2 produced.
Since the stoichiometry is 1:1, the number of moles of magnesium carbonate is equal to the number of moles of carbon dioxide.
Therefore, the number of moles of CO2 produced = 11.87 mol.
Step 3: Convert moles to volume using the ideal gas law.
The ideal gas law equation is PV = nRT, where:
P = pressure (1.00 atm)
V = volume of gas (to be determined)
n = number of moles of gas (11.87 mol)
R = ideal gas constant (0.08206 L·atm/(mol·K))
T = temperature in Kelvin (350 ˚C = 350 + 273.15 = 623.15 K)
Rearranging the equation:
V = nRT / P
= (11.87 mol) * (0.08206 L·atm/(mol·K)) * (623.15 K) / (1.00 atm)
≈ 582.54 L
Therefore, the volume of carbon dioxide produced at 350˚C and 1.00 atm from the complete thermal decomposition of 1.00 kg of magnesium carbonate is approximately 582.54 liters.
To find the density of carbon dioxide at this temperature, we can use the ideal gas law again.
Step 4: Calculate the density of carbon dioxide.
Density = mass / volume
The molar mass of CO2 = 12.01 g/mol (C) + 2 * 16.00 g/mol (O)
= 44.01 g/mol
Mass of CO2 = number of moles of CO2 * molar mass of CO2
= 11.87 mol * 44.01 g/mol
≈ 522.85 g
Converting g to kg:
Mass of CO2 = 522.85 g / 1000
≈ 0.52285 kg
Density = mass / volume
= 0.52285 kg / 582.54 L
≈ 0.000898 g/L
Therefore, the density of carbon dioxide at 350˚C is approximately 0.000898 g/L.
To find the volume of carbon dioxide produced, we need to use the ideal gas law equation, which states:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/K.mol)
T = temperature (in Kelvin)
First, let's convert the given temperature of 350˚C to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 350 + 273.15
T(K) = 623.15 K
Next, we need to find the number of moles of carbon dioxide produced from the decomposition of 1.00 kg of magnesium carbonate, MgCO3. To do this, we need to calculate the molar mass of MgCO3, and then convert the mass of MgCO3 to moles.
The molar mass of MgCO3 can be found by adding up the atomic masses of its constituent elements:
Mg: 24.31 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (3 atoms)
Molar mass of MgCO3 = (1 × 24.31) + (1 × 12.01) + (3 × 16.00)
Molar mass of MgCO3 = 24.31 + 12.01 + 48.00
Molar mass of MgCO3 = 84.32 g/mol
Now we can calculate the number of moles of MgCO3:
moles of MgCO3 = mass of MgCO3 / molar mass of MgCO3
moles of MgCO3 = 1.00 kg / 84.32 g/mol
moles of MgCO3 = 11.86 mol
Since magnesium carbonate decomposes into one mole of CO2, we can say that the number of moles of CO2 produced is also 11.86 moles.
Now we can calculate the volume of carbon dioxide using the ideal gas law equation (PV = nRT):
V = (nRT) / P
V = (11.86 mol × 0.0821 L.atm/K.mol × 623.15 K) / 1.00 atm
V = 616.13 L
Therefore, the volume of carbon dioxide produced is approximately 616.13 liters at a temperature of 350˚C and a pressure of 1.00 atm.
To calculate the density of carbon dioxide at this temperature, we need to know the molar mass of CO2. The molar mass of CO2 is:
C: 12.01 g/mol
O: 16.00 g/mol (2 atoms)
Molar mass of CO2 = (1 × 12.01) + (2 × 16.00)
Molar mass of CO2 = 12.01 + 32.00
Molar mass of CO2 = 44.01 g/mol
Now we can calculate the density of CO2 using the formula:
density = mass / volume
Since we have 11.86 moles of CO2, the mass of CO2 can be calculated as:
mass of CO2 = number of moles × molar mass
mass of CO2 = 11.86 mol × 44.01 g/mol
mass of CO2 = 521.91 g
Finally, we can calculate the density:
density = mass of CO2 / volume of CO2
density = 521.91 g / 616.13 L
density = 0.847 g/L
Therefore, the density of carbon dioxide at a temperature of 350˚C and a pressure of 1.00 atm is approximately 0.847 g/L.