Question

Gaseous ethane CH3CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 1.50 g of ethane is mixed with 3.7 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.

Answer 1

Balanced chemical reaction is

2CH₃CH₃ + 7O₂    4CO₂ + 6H₂O

Calculate limiting reactant

mass of ethane = 1.50 gm

mass of oxygen = 3.7 gm

molar mass of ethane = 30.07 g/mol

molar mass of O₂ = 31.9988 g/mol

no. of mole = gm of compound / molar mass

mole of ethane = 1.50 / 30.07 = 0.049884 mole

mole of O₂ = 3.7 / 31.9988 = 0.11563 moles

According to balanced chemical eqation 2 mole ethane react with  7 mole O₂​​​​​​​ molar ratio between thane to O₂​​​​​​​ is 2: 7 therefore to react with 0.049884 mole of ethane​​​​​​​ required O₂​​​​​​​ = 0.049884 X 7 / 2 = 0.174594 mole but O₂​​​​​​​ given only 0.11563 mole therefore O₂​​​​​​​ is limiting reactant.

According to balanced reaction 7 mole O₂​​​​​​​ produce 4 mole CO₂ molar ratio between O₂​​​​​​​  to CO₂ is 7:4 therefore 0.11563 mole of O₂​​​​​​​ produce CO₂ = 0.11563 X 4 / 7 = 0.06607 mole of CO₂

gm of compound = no. of mole X molar mass

moles of CO₂ = 0.06607 moles

molar mass of CO₂ = 44.01 g/mol

gm of aspirin = 0.06607 X 44.01 = 2.9 gm

Maximum mass of carbon dioxide that could be produced = 2.9 gm