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What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?
- a)2.8 kg
- b)6.4 kg
- c)9.6 kg
- d)96 kg
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
What is the weight of oxygen required for the complete combustion of 2...
C2H4 + 3 O2 —→ 2CO2 + 2H2O 28 kg 96 kg
∵ 28 kg of C2H4 undergo complete combustion by = 96 kg of O2
∴ 2.8 kg of C2H4 undergo complete combustion by = 9.6 g of O2.
∵ 28 kg of C2H4 undergo complete combustion by = 96 kg of O2
∴ 2.8 kg of C2H4 undergo complete combustion by = 9.6 g of O2.
Most Upvoted Answer
What is the weight of oxygen required for the complete combustion of 2...
Given:
Mass of ethylene = 2.8 kg
To find:
Weight of oxygen required for complete combustion of ethylene
Solution:
The combustion reaction of ethylene is given as:
C2H4 + 3O2 → 2CO2 + 2H2O
From the balanced equation, we can see that 1 mole of ethylene reacts with 3 moles of oxygen gas to produce 2 moles of carbon dioxide and 2 moles of water vapor.
The molar mass of ethylene (C2H4) is:
C = 12.01 g/mol
H = 1.01 g/mol
Therefore, the molar mass of ethylene = 2(12.01 g/mol) + 4(1.01 g/mol) = 28.05 g/mol
The number of moles of ethylene in 2.8 kg can be calculated as:
Number of moles of ethylene = mass of ethylene / molar mass of ethylene
= 2800 g / 28.05 g/mol
= 99.64 mol
From the balanced equation, we know that 1 mole of ethylene requires 3 moles of oxygen gas for complete combustion. Therefore, the number of moles of oxygen required for complete combustion of 99.64 moles of ethylene is:
Number of moles of oxygen = 3 x 99.64 = 298.92 mol
The molar mass of oxygen (O2) is 32 g/mol. Therefore, the mass of oxygen required for complete combustion of 2.8 kg of ethylene is:
Mass of oxygen = number of moles x molar mass
= 298.92 mol x 32 g/mol
= 9565.44 g
= 9.57 kg (rounded off to two decimal places)
Therefore, the weight of oxygen required for the complete combustion of 2.8 kg of ethylene is 9.6 kg.
Hence, the correct option is (c) 9.6 kg.
Mass of ethylene = 2.8 kg
To find:
Weight of oxygen required for complete combustion of ethylene
Solution:
The combustion reaction of ethylene is given as:
C2H4 + 3O2 → 2CO2 + 2H2O
From the balanced equation, we can see that 1 mole of ethylene reacts with 3 moles of oxygen gas to produce 2 moles of carbon dioxide and 2 moles of water vapor.
The molar mass of ethylene (C2H4) is:
C = 12.01 g/mol
H = 1.01 g/mol
Therefore, the molar mass of ethylene = 2(12.01 g/mol) + 4(1.01 g/mol) = 28.05 g/mol
The number of moles of ethylene in 2.8 kg can be calculated as:
Number of moles of ethylene = mass of ethylene / molar mass of ethylene
= 2800 g / 28.05 g/mol
= 99.64 mol
From the balanced equation, we know that 1 mole of ethylene requires 3 moles of oxygen gas for complete combustion. Therefore, the number of moles of oxygen required for complete combustion of 99.64 moles of ethylene is:
Number of moles of oxygen = 3 x 99.64 = 298.92 mol
The molar mass of oxygen (O2) is 32 g/mol. Therefore, the mass of oxygen required for complete combustion of 2.8 kg of ethylene is:
Mass of oxygen = number of moles x molar mass
= 298.92 mol x 32 g/mol
= 9565.44 g
= 9.57 kg (rounded off to two decimal places)
Therefore, the weight of oxygen required for the complete combustion of 2.8 kg of ethylene is 9.6 kg.
Hence, the correct option is (c) 9.6 kg.
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What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?a)2.8 kgb)6.4 kgc)9.6 kgd)96 kgCorrect answer is option 'C'. Can you explain this answer?
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