Gaseous ethane CH3CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. What is the theoretical yield of carbon dioxide formed from the reaction of 1.8g of ethane and 4.6g of oxygen gas?

mass of CO₂ produced = 5.06 g of CO₂

Explanation:

Equation of the reaction: 2CH₃CH₃ + 5O₂ ---> 4CO₂ + 6H₂O

From the equation of the reaction, 2 moles of gaseous ethane reacts with 5 moles of oxygen gas to produce 6 moles of water.

molar mass of ethane = 30 g/mol

molar mass of oxygen gas = 32 g/mol

number of moles of ethane present in 1.8 g = 1.8/30 = 0.06 moles

number of moles of oxygen gas present in 4.6 g = 4.6/32 = 0.14375 moles

mole ratio of oxygen gas to ethane  = 2.4 : 1

Therefore, oxygen is the limiting reactant

0.14375 moles of oxygen will react with 0.06 moles of ethane to produce 4/5 * 0.14375 moles of CO₂ = 0.115 moles of CO₂

molar mass of CO₂ = 44 g/mol

mass of CO₂ produced = 0.1725 * 44

mass of CO₂ produced = 5.06 g of CO₂

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2