Volume at NTP of oxygen required to completely burn 1 kg of coal
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Answered by
35
equation:
C + O2 → CO2
1mol C will react with 1 mol O2
Mass ratio 12:32
1kg coal will require 1*32/12 = 2.667kg = 2,667 g O2
Molar mass O2 = 32g/mol
2,667g = 2667/32 = 83.34mol
Ar NTP 1 mol = 24L
83.34 mol = 83.34*24 = 2000L
You ask about NTP = my understanding is that at NTP 1 mol gas = 24L
But at STP , 1 mol gas = 22.4L
At STP , 83.34mol = 83.34*22.4 = 1,867L OR 1.86*10^3 L
Volume of O2 required art STP = 1.86*10^3 L
C + O2 → CO2
1mol C will react with 1 mol O2
Mass ratio 12:32
1kg coal will require 1*32/12 = 2.667kg = 2,667 g O2
Molar mass O2 = 32g/mol
2,667g = 2667/32 = 83.34mol
Ar NTP 1 mol = 24L
83.34 mol = 83.34*24 = 2000L
You ask about NTP = my understanding is that at NTP 1 mol gas = 24L
But at STP , 1 mol gas = 22.4L
At STP , 83.34mol = 83.34*22.4 = 1,867L OR 1.86*10^3 L
Volume of O2 required art STP = 1.86*10^3 L
Answered by
4
Answer:
Mass of C = 1000 g
1 moles C requires 1 mole O₂ to burn.
Thus, 83.34 mole C requires 83.34 moles O₂
At NTP, 1 mole gas occupies 22.4 litres.
83.34 moles occupies 83.34 × 22.4 = 1.86 × 10³ L
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Explanation:
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