50cm³ of methane CH4 was exploded with 170cm³ of oxygen gas and underwent complete combustion
CH4+2O2~CO2+2H2O
Determine the total volume of the resultant gaseous mixture and the volume of each gas
To determine the total volume of the resultant gaseous mixture, we need to calculate the moles of methane, moles of oxygen, moles of carbon dioxide, and moles of water produced.
1. Moles of methane:
Using the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Since we have the volume and T is constant, we can simplify the equation to n = PV/RT.
Given that the volume of methane is 50 cm³, we need to convert it to liters: 50 cm³ = 0.05 liters.
Assuming the temperature is constant, we can use the ideal gas constant R = 0.0821 L·atm/(mol·K).
Now, we can calculate the moles of methane:
n(CH4) = (0.05 L)/(0.0821 L·atm/(mol·K)) ≈ 0.61 moles
2. Moles of oxygen:
According to the balanced chemical equation, the ratio of methane to oxygen is 1:2. Therefore, we need twice as many moles of oxygen as moles of methane:
n(O2) = 2 * 0.61 moles ≈ 1.22 moles
3. Moles of carbon dioxide:
From the balanced chemical equation, we see that for every mole of methane, one mole of carbon dioxide is produced.
Thus, the number of moles of carbon dioxide is the same as the number of moles of methane:
n(CO2) = 0.61 moles
4. Moles of water:
According to the balanced chemical equation, for every mole of methane, two moles of water are produced.
Thus, the number of moles of water is twice the number of moles of methane:
n(H2O) = 2 * 0.61 moles ≈ 1.22 moles
Now that we have determined the moles of all the gases, we can calculate their respective volumes using the ideal gas law.
For methane:
V(CH4) = n(CH4) * RT/P
Using P = 1 atm and T = 273 K (assuming standard temperature and pressure),
V(CH4) = (0.61 mol) * (0.0821 L·atm/(mol·K)) * (273 K) / (1 atm) ≈ 13.5 liters
For oxygen:
V(O2) = n(O2) * RT/P
V(O2) = (1.22 mol) * (0.0821 L·atm/(mol·K)) * (273 K) / (1 atm) ≈ 26.9 liters
For carbon dioxide:
V(CO2) = n(CO2) * RT/P
V(CO2) = (0.61 mol) * (0.0821 L·atm/(mol·K)) * (273 K) / (1 atm) ≈ 13.5 liters
For water:
V(H2O) = n(H2O) * RT/P
V(H2O) = (1.22 mol) * (0.0821 L·atm/(mol·K)) * (273 K) / (1 atm) ≈ 26.9 liters
Lastly, to find the total volume of the resultant gaseous mixture, we add the volumes of all the gases:
Total volume = V(CH4) + V(O2) + V(CO2) + V(H2O)
Total volume ≈ 13.5 L + 26.9 L + 13.5 L + 26.9 L = 80.8 liters
Therefore, the total volume of the resultant gaseous mixture is approximately 80.8 liters. The volume of methane is approximately 13.5 liters, the volume of oxygen is approximately 26.9 liters, the volume of carbon dioxide is approximately 13.5 liters, and the volume of water is approximately 26.9 liters.