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Ray Of Light [21]
3 years ago
7

Gaseous methane CH4 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 12.4g of c

arbon 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.
Chemistry
1 answer:
k0ka [10]3 years ago
8 0

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 %

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The question is incomplete, complete question is :

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