(a.) The balanced reaction is : 2 C4H10 + 13 O2 8 CO2 + 10 H2O
(b.) Case 1 : butane is the limiting reactant
mass C4H10 = 23.2 g
moles C4H10 = (mass C4H10) / (molar mass C4H10)
moles C4H10 = (23.2 g) / (58.12 g/mol)
moles C4H10 = 0.400 mol
moles CO2 = (moles C4H10) * (8 moles CO2 / 2 moles C4H10)
moles CO2 = (0.400 mol) * (8 / 2)
moles CO2 = (0.400 mol) * (4)
moles CO2 = 1.60 mol
Case 2 : oxygen is the limiting reactant
mass O2 = 93.7 g
moles O2 = (mass O2) / (molar mass O2)
moles O2 = (93.7 g) / (32.0 g/mol)
moles O2 = 2.93 mol
moles CO2 formed = (moles O2) * (8 moles CO2 / 13 moles O2)
moles CO2 formed = (2.93 mol) * (8/13)
moles CO2 formed = 1.80 mol
Since less moles of CO2 are formed in Case 1, therefore, butane is the limiting reactant.
(c.) Theoretical moles CO2 formed = 1.60 mol
Theoretical yield CO2 = (Theoretical moles CO2 formed) * (molar mass CO2)
Theoretical yield CO2 = (1.60 mol) * (44.0 g/mol)
Theoretical yield CO2 = 70.4 g
(d.) Percent yield = (actual yield / theoretical yield) * 100
Percent yield = (69.2 g / 70.4 g) * 100
Percent yield = (0.983) * 100
Percent yield = 98.3 %
If 23.2 grams of butane (CaH10) and 93.7 grams of oxygen (O2) are available in the...
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