Question

Calculate the mass of carbon(IV)oxide produced by burning 150 grams of butane (C4H10)

Answer 1

2C4H10 + 13O2 ==> 8CO2 + 10H2O

moles butane = g/molar mass = 150/58 = 2.59
mols CO2 produced = 2.59 mols C4H10 x (8 mols CO2/2 mols C4H10) = 2.59 x 8/2 = 2.59 x 4 = 10.4
Then grams = mols CO2 x molar mass CO2 = ?