PV = nRT and solve for n = number of mols.
Then n = grams/molar mass. You know molar mass and n, solve for grams.
Then n = grams/molar mass. You know molar mass and n, solve for grams.
PV = nRT
Where:
P = pressure of the gas (in Pascals)
V = volume of the gas (in cubic meters)
n = number of moles of gas
R = ideal gas constant (8.314 J/(mol·K))
T = temperature of the gas (in Kelvin)
First, let's convert the given values to appropriate units:
Pressure (P) = 105 kPa = 105,000 Pa
Volume (V) = 20 dm3 = 0.02 m3 (since there are 1000 dm3 in 1 m3)
Temperature (T) = 20 °C = 20 + 273 = 293 K
Now, rearrange the ideal gas law equation to calculate the number of moles (n) of carbon dioxide:
n = PV / RT
n = (105,000 Pa) x (0.02 m3) / (8.314 J/(mol·K) x 293 K)
Calculate the value of n:
n ≈ 0.851 mol
Now, we need to determine the molar mass of carbon dioxide (CO2), which can be calculated by adding the atomic masses of carbon (C) and two oxygen (O) atoms.
C = 12.01 g/mol
O = 16.00 g/mol
Molar mass of CO2 = 12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol
Finally, to find the mass of carbon dioxide produced (m), we can use the equation:
m = n x M
m = 0.851 mol x 44.01 g/mol
Calculate the value of m:
m ≈ 37.42 g
Therefore, approximately 37.42 grams of carbon dioxide were produced.