When using a limited quantity of oxygen to react with propane (C3H8), this will result in producing carbon monoxide (CO) and carbon dioxide (CO2) gases. If 25% of 128 g of propane used in such process is converted into CO and the rest into CO2, calculate:
1- the volume of oxygen used at STP.
2- the mass of water vapor produced in such process.
Given that:
Atomic mass C=12, O=16, H=1
C3H8 + O2 = CO + H2O (Unbalanced)
Thank you!
To solve this problem, we will first balance the chemical equation, then calculate the moles of propane consumed, moles of CO formed, and moles of CO2 formed. From the balanced equation, we can determine the moles of oxygen consumed, and finally, use the ideal gas law to calculate the volume of oxygen at STP.
1. Balancing the equation:
C3H8 + 5O2 → 3CO2 + 4H2O
Now, we can see that 3 moles of CO2 are formed per mole of C3H8, and 4 moles of H2O are formed per mole of C3H8.
2. Calculate the moles of propane consumed:
Given that the molar mass of propane (C3H8) is:
(3 x atomic mass of C) + (8 x atomic mass of H) = (3 x 12) + (8 x 1) = 36 + 8 = 44 g/mol
So, the moles of propane used can be calculated as:
Number of moles = Mass / Molar mass
Number of moles = 128 g / 44 g/mol = 2.909 mol
3. Calculate the moles of CO and CO2 formed:
Since 25% of the propane is converted into CO, and the remaining propane is converted into CO2, we can calculate the moles of CO and CO2 formed as follows:
Moles of CO = (25 / 100) x 2.909 mol = 0.727 mol
Moles of CO2 = (75 / 100) x 2.909 mol = 2.181 mol
4. Calculate the moles of oxygen consumed:
From the balanced equation, we see that 1 mole of propane reacts with 5 moles of oxygen. So, the moles of oxygen consumed can be calculated as:
Moles of oxygen = 5 x moles of propane = 5 x 2.909 mol = 14.545 mol
5. Calculate the volume of oxygen at STP:
At STP (Standard Temperature and Pressure), the volume of 1 mole of any ideal gas is 22.4 L.
So, the volume of oxygen used can be calculated as:
Volume of oxygen = Moles of oxygen x Molar volume at STP
Volume of oxygen = 14.545 mol x 22.4 L/mol = 325.856 L
Therefore, the volume of oxygen used at STP is approximately 325.856 L.
6. Calculate the mass of water vapor produced:
From the balanced equation, we can see that 4 moles of H2O are formed per mole of C3H8 consumed. So, the moles of water vapor can be calculated as:
Moles of water vapor = 4 x moles of propane = 4 x 2.909 mol = 11.636 mol
Given that the molar mass of water (H2O) is:
(2 x atomic mass of H) + (1 x atomic mass of O) = (2 x 1) + (1 x 16) = 2 + 16 = 18 g/mol
The mass of water vapor produced can be calculated as:
Mass of water vapor = Moles of water vapor x Molar mass
Mass of water vapor = 11.636 mol x 18 g/mol = 209.448 g
Therefore, the mass of water vapor produced is approximately 209.448 g.
To begin, let's balance the chemical equation:
C3H8 + 5O2 = 3CO2 + 4H2O
Now, let's calculate the moles of propane (C3H8) used in the reaction:
Molar mass of C3H8 = (3 * 12.01 g/mol) + (8 * 1.01 g/mol) = 44.1 g/mol
Moles of C3H8 = 128 g / 44.1 g/mol ≈ 2.90 mol
According to the reaction, for every 1 mole of propane, 3 moles of carbon dioxide (CO2) are produced. Therefore, the moles of carbon dioxide produced in the reaction can be calculated as:
Moles of CO2 = 3 * 2.90 mol = 8.70 mol
Since 25% of the propane is converted into carbon monoxide (CO), the moles of carbon monoxide produced can be calculated as:
Moles of CO = (25/100) * 2.90 mol = 0.73 mol
Now let's find the moles of oxygen (O2) used in the reaction. From the balanced chemical equation, we know that for every 5 moles of oxygen, 1 mole of propane is used. Therefore:
Moles of O2 = (5/1) * 2.90 mol = 14.5 mol
To calculate the volume of oxygen used at STP, we can use the ideal gas law equation:
PV = nRT
At STP (Standard Temperature and Pressure), the temperature (T) is 273.15 K and the pressure (P) is 1 atm. The gas constant (R) is 0.0821 L.atm/(mol.K).
V = (nRT)/P = (14.5 mol * 0.0821 L.atm/(mol.K) * 273.15 K) / 1 atm ≈ 326 L
Therefore, the volume of oxygen used at STP is approximately 326 liters.
Now, let's calculate the mass of water vapor (H2O) produced in the process:
From the balanced chemical equation, we know that for every 4 moles of water vapor, 1 mole of propane is used. Therefore, the moles of water vapor produced can be calculated as:
Moles of H2O = (4/1) * 2.90 mol = 11.60 mol
Molar mass of H2O = (2 * 1.01 g/mol) + (16.00 g/mol) = 18.02 g/mol
Mass of H2O = Moles of H2O * Molar mass of H2O = 11.60 mol * 18.02 g/mol ≈ 209.22 g
Therefore, the mass of water vapor produced in the process is approximately 209.22 grams.