Question

I would really appreciate it if you could explain this problem step by step, in as much detail as possible, as I am having a difficult time trying to understand it. Thank you :)

How many liters of CO2 can be produced by combustion of 5.80 g butane with 15.0 liters of oxygen . All gases measured at STP.

I appreciate your help!

Answer 1

balanced reaction is

2C₄H₁₀  + 13O₂   8CO₂ + 10H₂O

According to reaction 2 mole of butane react with 13 mole of oxygen 2 produce 8 mole of CO₂

first calculate mole of butane and mole of oxygen

molar mass of butane = 58.12 gm/mol then 5.80 gm = 5.80/ 58.12 = 0.09979 mole of butane

At STP 1mole of gas occupy volume 22.414 L then 15 L oxygen = 15/22.414 = 0.6692 mole

According to reaction for complete reaction 2 mole of butane require 13 mole of oxygen then for 0.09979 mole of butane require 0.09979 13 / 2 = 0.6486 mole of oxygen but oxygen given 0.6692 mole thus oxygen is excess reactant and butane is limiting reactant react completly

According to reaction 2 mole of butane produce 8 mole of CO₂ then 0.09979 mole of butane produce

0.09979 8 / 2 = 0.39916 mole of CO₂

1 mole of gas at STP occupy volume 22.414 L then 0.39916 mole of CO₂ occupy volume

22.414 0.39916 = 8.95 L

8.95 L liters of CO₂ can be produced by combustion of 5.80 g butane with 15.0 liters of oxygen