Question

44 gram sample of a natural gas consisting of Methane and ethylene was burnt in excess of oxygen yielding 132 gram of carbon dioxide and some water as product what is the mole percentage of Ethylene in the sample

Answer 1

Given:

44 gram sample of a natural gas consisting of Methane and ethylene was burnt in excess of oxygen yielding 132 gram of carbon dioxide and some water as product

To find:

What is the mole percentage of Ethylene in the sample

Solution:

From given, we have,

The reaction is given by the equation:

CH₄ + C₂H₄ + 5O₂ → 3CO₂+4H₂O

16 g of CH4 gives 132 g of CO2

1 mol of C2H4 gives 3 mol of CO2

28 g of C2H4 gives 132 g of CO2

44 gram sample of a natural gas

Moles of CO2 formed in the reaction = 132/44 = 3  mol

From the reaction, we know that 3 mol of CO2 is formed by 1 mol of CH4

Therefore, Moles of CH4 = 1 and  the mass of CH4 = 16 g

44 gram sample of a natural gas

Therefore, the mass of C2H4 = 44 − 16 = 28 g = 1 mol of ethylene

Mole % of ethylene = Moles of ethylene/Total moles × 100 = 1/2 × 100 = 50%

Therefore, 50 % of Ethylene is in the sample

Answer 2

Answer:

50%

Explanation:

Reaction =

CH4 + C2H4 + 5O2 -----> 3CO2 + 4H20

moles of CO2 = mass ÷ molar mass

moles of CO2 = 132 gm ÷ 44

moles of CO2 = 3 moles

1 mole of C2H4 give 3 moles of CO2

1 mole of CH4 give 3 moles of CO2

so,

Total Moles in Mixture = 1 + 1

Total Moles in Mixture = 2 moles

Moles of C2H4 = 1 mole

Mole % of C2H4 = (1 ÷ 2)× 100

Mole % of C2H4 = 50%