Question

Gaseous methane CH4 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 12.4g of carbon dioxide is produced from the reaction of 7.38g of methane and 39.2g of oxygen gas, calculate the percent yield of carbon dioxide. Round your answer to 3 significant figures.

Answer 1

Answer:

The percent yield = 61.8 %

Explanation:

Step 1: Data given

Mass of CO2 produced = 12.4 grams

Mass of methane = 7.38 grams

Mass of oxygen = 39.2 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of methane = 16.04 g/mol

Molar mass of oxygen = 32.0 g/mol

Step 2: The balanced equation

CH4 + 2O2 → CO2 + 2H2O

Step 3: Calculate moles methane

Moles methane = mass methane / molar mass methane

Moles methane = 7.38 grams / 16.04 g/mol

Moles methane = 0.460 moles

Step 4: Calculate moles oxygen

Moles O2 = 39.2 grams / 32.0 g/mol

Moles O2 = 1.225 moles

Step 5: Calculate limiting reactant

For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

Methane is the limiting reactant. It will completely be consumed (0.460 moles). O2 is in excess. There will react 2*0.460 = 0.920 moles

There will remain 1.225 -0 .920 = 0.305 moles O2

Step 6: Calculate moles of CO2

For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

For 0.460 moles CH4 we'll have 0.460 moles CO2

Step 7: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.460 moles * 44.01 g/mol

Mass CO2 = 20.24 grams

Step 8: Calculate percent yield

% yield = (actual yield/theoretical yield) *100%

% yield = (12.5 grams / 20.24 grams ) *100%

% yield = 61.8 %

The percent yield = 61.8 %