Question

Volume of oxygen at NTP, required to completely burn 1 kg of coal $(100\%$ arbon) is ___________.

• A. 22.4 L
• B. $1.86\times {10}^3$ L
• C. $22.4\times {10}^3$ L
• D. 1000 L

Answer 1

The correct option is B $1.86\times {10}^3$ 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