Gaseous methane (CH4) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2o) . What is the theoretical yield of carbon dioxide formed from the reaction of 0.16g of methane and 0.83g of oxygen gas?
CH4+2O2>> CO2 + 2H2O
so for each mole of methane, you get one mole of carbon dioxide.
So we have to see the limiting agents.
You are given .16/16 moles .01 moles of methane. You must have twice that of oxygen see the ratio moles O2=.83/32 which is greater than .02, so the limiting reagent is methane. Then you ought to get .01 moles carbon dioxide, ord .44 grams.
To find the theoretical yield of carbon dioxide formed from the reaction of methane and oxygen gas, we need to follow these steps:
1. Determine the balanced chemical equation for the reaction:
CH4 + 2O2 -> CO2 + 2H2O
2. Calculate the molar masses of methane (CH4) and oxygen gas (O2):
molar mass of CH4 = (1 x atomic mass of C) + (4 x atomic mass of H) = 12.01 g/mol + (4 x 1.01 g/mol) = 16.05 g/mol
molar mass of O2 = (2 x atomic mass of O) = (2 x 16.00 g/mol) = 32.00 g/mol
3. Convert the given masses of methane and oxygen gas into moles:
moles of CH4 = mass of CH4 / molar mass of CH4 = 0.16g / 16.05 g/mol = 0.00996 mol
moles of O2 = mass of O2 / molar mass of O2 = 0.83g / 32.00 g/mol = 0.02594 mol
4. Determine the limiting reactant:
The limiting reactant is the one that is completely consumed first and limits the amount of product formed. To find the limiting reactant, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation. From the balanced equation, we can see that 1 mole of CH4 reacts with 2 moles of O2.
moles of O2 required = 2 * moles of CH4 = 2 * 0.00996 = 0.01992 mol
Since we have 0.02594 mol of O2, it is in excess, and methane (CH4) is the limiting reactant.
5. Calculate the moles of CO2 formed using the stoichiometry of the balanced equation:
From the balanced equation, we can see that 1 mole of CH4 produces 1 mole of CO2.
moles of CO2 = moles of CH4 = 0.00996 mol
6. Calculate the mass of CO2 formed using the molar mass of CO2:
mass of CO2 = moles of CO2 * molar mass of CO2 = 0.00996 mol * 44.01 g/mol = 0.439 g
Therefore, the theoretical yield of carbon dioxide formed from the reaction of 0.16g of methane and 0.83g of oxygen gas is 0.439 grams.
To find the theoretical yield of carbon dioxide (CO2) formed from the reaction of methane (CH4) and oxygen gas (O2), we need to determine which reactant is the limiting reagent.
Step 1: Convert the given masses of methane and oxygen gas to moles.
- Moles of methane (CH4):
0.16g CH4 * (1 mol CH4/16.04 g CH4) ≈ 0.00998 mol CH4
- Moles of oxygen gas (O2):
0.83g O2 * (1 mol O2/32.00 g O2) ≈ 0.0259 mol O2
Step 2: Write a balanced chemical equation for the reaction.
CH4 + 2O2 → CO2 + H2O
Step 3: Determine the stoichiometry of the reactants and products in the balanced equation.
From the balanced equation, it shows that 1 mole of CH4 produces 1 mole of CO2.
Step 4: Determine the number of moles of carbon dioxide (CO2) produced.
Since 1 mole of CH4 produces 1 mole of CO2, the moles of CO2 produced is also 0.00998 mol.
Step 5: Convert moles of carbon dioxide (CO2) to grams.
Moles of CO2 = 0.00998 mol CO2
Molar mass of CO2 = 44.01 g/mol
Mass of CO2 = 0.00998 mol * 44.01 g/mol ≈ 0.439 g CO2
Therefore, the theoretical yield of carbon dioxide formed from the reaction of 0.16 g of methane and 0.83 g of oxygen gas is approximately 0.439 g CO2.