A mole of methane reacts with 64g of oxygen at 1 atm, 425 K. There are two ways in which methane can react with oxygen. It can either form water vapor and carbon dioxide or water vapor and carbon monoxide. After reaction the final gas density is 0.7282 g/L.
What is the fraction of methane that reacted to produce CO2? What is the fraction of methane that reacted to produce CO?
(Assume that methane is fully reacted)
Can someone just guide me in the right direction?
Here's what I have so far:
64g O2 --> 2 mol O2
CH4 + 2 O2 --> 2 H2O + CO2
2 CH4 + 3 O2 --> 4 H20 + 2 CO
To determine the fraction of methane that reacted to produce CO2 and CO, we need to consider the balanced chemical equation and the stoichiometry of the reaction.
From the balanced equation, we know that for every 2 moles of methane (CH4), we get 1 mole of carbon dioxide (CO2) and for every 2 moles of methane, we get 2 moles of water (H2O). Similarly, for every 2 moles of methane, we get 1 mole of carbon monoxide (CO) and 4 moles of water (H2O).
First, we need to calculate the number of moles of methane used in the reaction:
Given that 1 mole of methane reacts with 2 moles of oxygen, the number of moles of oxygen used is (64g O2) / (32 g/mol O2) = 2 moles of oxygen.
Since there is a 1:2 ratio of methane to oxygen in the reaction, the number of moles of methane used is (1/2) * 2 = 1 mole of methane.
Now, let's calculate the moles of carbon dioxide and carbon monoxide produced:
From the balanced equation, the ratio of moles of methane reacted to moles of carbon dioxide formed is 2:1. Therefore, the number of moles of carbon dioxide produced is (1/2) * 1 = 0.5 moles.
Similarly, the ratio of moles of methane reacted to moles of carbon monoxide formed is 2:1. Therefore, the number of moles of carbon monoxide produced is (1/2) * 1 = 0.5 moles.
Next, we need to calculate the volume of the final gas in liters:
The ideal gas law equation can be used to calculate the volume of gas:
PV = nRT
Given that the pressure (P) is 1 atm, the temperature (T) is 425 K, and the density (d) is 0.7282 g/L, we can rearrange the ideal gas law equation to solve for volume (V):
V = (molar mass / density) * (RT)
The molar mass of the final gas can be calculated by adding the molar masses of carbon dioxide (44 g/mol) and carbon monoxide (28 g/mol):
Molar mass of the final gas = (0.5 moles * 44 g/mol) + (0.5 moles * 28 g/mol) = 36 g/mol
Using the given values for pressure, temperature, density, and the molar mass of the final gas, we can calculate the volume (V).
Finally, to determine the fraction of methane that reacted to produce CO2 and CO:
Fraction of methane reacted to produce CO2 = (0.5 moles CO2 * molar mass of CO2) / (1 mole CH4 * molar mass of CH4)
Fraction of methane reacted to produce CO = (0.5 moles CO * molar mass of CO) / (1 mole CH4 * molar mass of CH4)
By plugging in the respective molar masses and solving, you will obtain the fraction of methane reacted to produce CO2 and CO.