Question

Volume of ntp of oxygen required to completely burn 1 kg of coal 100% carbon

Answer 1

Answer:

Explanation:

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

Answer 2

Answer:

C + O2 → CO2

1mol C will react with 1 mol O2

Mass ratio 12:32

1kg coal will require 1*32/12 = 2.667kg = 2667 g O2

Molar mass O2 = 32g/mol

2,667g = 2667/32 = 83.34mol

At NTP 1 mol = 24L

83.34 mol = 83.34*24 = 2000L

at NTP 1 mol gas = 24L

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

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Explanation: