Question

Gaseous butane (CH3(CH2)2 CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2)...

Gaseous butane (CH3(CH2)2 CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water ( H2O). If 67.2 g of carbon dioxide is produced from the reaction of 38.94 g of butane and 202.1 g of oxygen gas, Calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.

Homework Answers

Answer #1

Balanced equation for the reaction:

2C₄H₁₀ + 13O₂ --> 8CO₂ + 10H₂O

Given: 38.94 g of butane & 202.1 g of oxygen gas

67.2 g of carbon dioxide produced

According to balanced equation

For 1 mol of butane 6.5 mols of oxygen is required and it yields in 4 moles of CO2 and 5 moles of H2O.

38.94 g of butane is 0.67 moles and 202.1 of oxygen is 6.3 moles

but 0.67 moles of butane requires only 4.4 moles of oxygen.

Therefore limiting reagent is butane

Therefore 1 mole butane --------àgives 4 moles of CO2

                 0.67 moles of butane -------à ? moles of CO2 [X]

X = 0.67 X 4/1 = 2.68 moles of CO2 is produced = 2.68 X 44 = 117.9 g of CO2 should be produced but we have got only 67.2 g = 67.2/44 = 1.53 moles

Percent yield = actual yield/theoretical yield X 100 = 56.9%

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