How many grams of carbon can be burned in 15 dm3 of oxygen and many dm3 of CO2 is produced?
To determine the number of grams of carbon that can be burned in 15 dm3 of oxygen, we first need to understand the balanced chemical equation for the combustion of carbon:
C + O2 -> CO2
This equation tells us that 1 mole of carbon (12 grams) reacts with 1 mole of oxygen (32 grams) to produce 1 mole of carbon dioxide (44 grams).
Next, we need to determine the number of moles of oxygen in 15 dm3. To do this, we need to use the ideal gas law, which states:
PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
Since we are given volume in dm3 and need to convert it to liters, we have:
V = 15 dm3 = 15 L
Assuming standard temperature and pressure (273 K and 1 atm), we can rearrange the ideal gas law to solve for n:
n = PV / RT
Substituting the given values:
n = (1 atm) * (15 L) / (0.0821 L.atm/mol.K * 273 K)
n ≈ 0.663 moles of oxygen
Since the molar ratio of oxygen to carbon dioxide is 1:1, we know that 0.663 moles of carbon dioxide will be produced.
Finally, we can convert the moles of carbon dioxide to grams using the molar mass of CO2:
44 grams CO2 / 1 mole CO2 * 0.663 moles CO2 ≈ 29.172 grams of CO2
Therefore, approximately 29.172 grams of carbon can be burned in 15 dm3 of oxygen, and approximately 29.172 dm3 of carbon dioxide is produced.