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# The combustion of butane is represented by the following equation:?

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

a) If 4.74 g of Butane react with excess oxygen, what volume of CO2, measured at 150. degrees Celsius and 1.14 atm, will be produced.

b) What mass of water is produced when butane burns and produces 3720 L of carbon dioxide, measured at 35 degrees Celsius and 985 kPa?

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- Roger the MoleLv 78 years agoFavorite Answer
a)

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

(4.74 g C4H10) / (58.12259 g C4H10/mol) x (8 mol CO2 / 2 mol C4H10) =

0.32621 mol CO2

PV = nRT

V = nRT / P = (0.32621 mol) x (0.08205746 L atm/K mol) x (150 + 273 K) /

(1.14 atm) = 9.93 L CO2

b) PV = nRT

n = PV / RT = (985 kPa) x (3720 L) / ((8.3144621 L kPa/K mol) x (35 + 273 K)) =

1430.85 mol CO2

(1430.85 mol CO2) x (10 mol H2O / 8 mol CO2) x (18.01532 g H2O/mol) = 32222 g =

32.2 kg H2O

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