Given the equation below, how many moles of butane (C4H10) must be burned in an excess of O2 to produce 150 g CO2?
2C4H10(g) + 13O2(g) --> 8CO2(g) + 10 H2O(g)
0.85 mol
To determine the number of moles of butane needed to produce 150 g of CO2, we need to follow these steps:
Step 1: Calculate the molar mass of CO2.
The molecular formula of CO2 consists of one carbon atom (mass = 12.01 g/mol) and two oxygen atoms (mass = 16.00 g/mol). So, the molar mass of CO2 is:
(1 * 12.01 g/mol) + (2 * 16.00 g/mol) = 44.01 g/mol
Step 2: Determine the moles of CO2.
To find the moles of CO2, we divide the mass of CO2 by its molar mass:
150 g / 44.01 g/mol = 3.40 moles of CO2
Step 3: Use the stoichiometry of the balanced equation to determine the moles of butane.
According to the balanced equation, 2 moles of C4H10 reacts to produce 8 moles of CO2. Therefore, for every 8 moles of CO2, we need 2 moles of C4H10.
So, to find the moles of butane:
(3.40 moles CO2 / 8 moles CO2) * 2 moles C4H10 = 0.85 moles of C4H10
Therefore, 0.85 moles of butane must be burned to produce 150 g of CO2.
To find out how many moles of butane (C4H10) must be burned to produce 150 g of CO2, we need to use stoichiometry.
1. Start by balancing the equation to ensure the number of atoms on both sides is equal:
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
2. Determine the molar mass of CO2.
Carbon (C) has a molar mass of 12.01 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol. Therefore, the molar mass of CO2 is:
(12.01 g/mol x 1) + (16.00 g/mol x 2) = 44.01 g/mol
3. Convert the mass of CO2 to moles.
Moles = Mass / Molar Mass
Moles of CO2 = 150 g / 44.01 g/mol ≈ 3.41 moles
4. Use the stoichiometric coefficients from the balanced equation to determine the moles of butane.
According to the balanced equation, the ratio of CO2 to C4H10 is 8:2. Therefore, the moles of C4H10 can be calculated as follows:
Moles of C4H10 = (Moles of CO2) x (2 moles of C4H10 / 8 moles of CO2)
Moles of C4H10 = 3.41 moles x (2 moles / 8 moles) = 0.8525 moles (rounded to four decimal places)
Therefore, approximately 0.8525 moles of butane (C4H10) must be burned in an excess of O2 to produce 150 g of CO2.
2C4H10(g) + 13O2(g) --> 8CO2(g) + 10 H2O(g)
mols CO2 needed = grams/molar mass = 150/44= 3.41
Convert mols CO2 to mols butane using the coefficients in the balanced equation like this.
3.41 mols CO2 x (2 mols butane/8 mols CO2) = ?