2CH3OH + 3O2→ 2CO2 + 4H2O

When 9.63 g of CH3OH was mixed with 15.7 g of O2 and ignited, 9.46 g of CO2 was obtained. What was the percentage yield of CO2?

Since when is 'north seattle college' a subject?

To calculate the percentage yield of CO2, we need to determine the theoretical yield (the amount of CO2 that would be produced if the reaction went to completion) and compare it to the actual yield (the amount of CO2 obtained in the experiment).

First, let's calculate the molar mass (in grams per mole) of each substance involved in the reaction:

- Molar mass of CH3OH (methanol) = (12.01 g/mol × 1) + (1.01 g/mol × 4) + (16.00 g/mol × 1) = 32.04 g/mol
- Molar mass of O2 = (16.00 g/mol × 2) = 32.00 g/mol
- Molar mass of CO2 = (12.01 g/mol × 1) + (16.00 g/mol × 2) = 44.01 g/mol

Next, we need to determine the moles of CH3OH and O2 used in the reaction. We can use the given masses and the molar masses to calculate the moles:

- Moles of CH3OH = Mass of CH3OH / Molar mass of CH3OH = 9.63 g / 32.04 g/mol = 0.3006 mol
- Moles of O2 = Mass of O2 / Molar mass of O2 = 15.7 g / 32.00 g/mol = 0.4906 mol

Now, we need to determine the stoichiometry of the reaction to find the mole ratio between CH3OH and CO2. The balanced chemical equation shows that 2 moles of CH3OH produce 2 moles of CO2:

2 CH3OH + 3 O2 → 2 CO2 + 4 H2O

Therefore, the moles of CO2 that would be produced when 0.3006 mol of CH3OH is used can be determined using the mole ratio:

Moles of CO2 = Moles of CH3OH = 0.3006 mol

We can now calculate the theoretical yield of CO2 by multiplying the moles of CO2 by the molar mass of CO2:

Theoretical yield of CO2 = Moles of CO2 × Molar mass of CO2 = 0.3006 mol × 44.01 g/mol = 13.21 g

Finally, we can calculate the percentage yield of CO2 using the formula:

Percentage yield = (Actual yield / Theoretical yield) × 100

Given that the actual yield of CO2 was 9.46 g, the percentage yield is calculated as:

Percentage yield = (9.46 g / 13.21 g) × 100 = 71.7%

Therefore, the percentage yield of CO2 in this reaction is approximately 71.7%.