Question

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.

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%