Question

Gaseous butane(CH₃(CH₂)₂CH₃) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dixoide (CO₂) and gaseous water(H₂O). What is the theoretical yield of water formed from the reaction of 36.6 g of butane and 203. g of oxygen gas? Be sure your answer has the correct number of significant digits in it.

Answer 1

Balanced chemical reaction is

2 CH₃CH₂CH₂CH₃  + 13 O₂    8CO₂  + 10H₂O

first calculate limiting reactant

molar mass of burane = 58.12 gm/mole then 2 mole of butane = 116.24 gm

molar mass of O₂ = 31.9988 gm/mole then 13 mole O₂ = 31.9988 13 = 415.9844

According to reaction 2 mole of butane react with 13 mole of O₂ that mean 116.24 gm of butane react with 415.9844 gm of O₂ then to react with 36.6 gm of butane required O₂ = 36.6 415.9844 / 116.24 = 130.98 gm

but O₂ given 203 gm thus O₂ is excess reactant and butane is limiting reactant.

molar mass of water = 18.01528 gm /mol then 10 mole of water = 180.1528 gm

According to reaction 2 mole of butane produce 10 mole of water that mean 116.24 gm butane produce 180.1528 gm of water then 36.6 gm butane produce = 36.6 180.1528 /116.24 = 56.724 gm of water

36.6 gm of butane produce 56.724 gm of water

theoretical yield = 56.724 gm water