Question

Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If of carbon dioxide is produced from the reaction of of methane and of oxygen gas, calculate the percent yield of carbon dioxide. Round your answer to significant figures.

Answer 1

The question is incomplete, the complete question is;

Gaseous methane (CH4) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 28.2 g of carbon dioxide is produced from the reaction of 15.1 g of methane and 81.2 g of oxygen gas, Calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.

Answer:

71.1%

Explanation:

The balanced reaction equation must first be written;

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

Let us obtain the number of moles of carbon dioxide corresponding to 28.2 g

Number of moles = mass/ molar mass= 28.2/44.01 g/mol = 0.64 moles of CO2

Next, we obtain the limiting reactant. This is the reactant that yields the least amount of product.

For methane;

Number of moles in 15.1 g= mass/molar mass= 15.1/16gmol-1 = 0.9 moles

From the reaction equation;

1 mole of methane yields 1 mole of carbon dioxide

Hence 0.9 moles of methane yields 0.9 moles of carbon dioxide.

For oxygen

Number of moles of oxygen corresponding to 81.2 g of oxygen= mass/ molar mass= 81.2g/32gmol-1 = 2.5 moles of oxygen

From the reaction equation;

2 moles of oxygen gas yields 1 mole of carbon dioxide

2.5 moles of oxygen gas yields 2.5 × 1 /2 = 1.25 moles of carbon dioxide.

Methane is the limiting reactant.

Theoretical yield of carbon dioxide= 0.9 moles

Actual yield of carbon dioxide= 0.64 moles

% yield = actual yield/ theoretical yield × 100

% yield= 0.64/0.9 ×100

% yield = 71.1%