volume at NTP of oxygen required to completely burn 1 Kg of coal (100% carbon )
solve it ?

Reaction for complete combustion is as follows:
C   +   O2    →   CO2
1 mole of carbon or 12 g of carbon require 1 mole of Oxygen or 32 g of Oxygen
1000 g or 1 kg of Carbon will react with 3212×1000
= 2666.67 g of Oxygen
1 mole of  oxygen or 32 g of Oxygen at NTP occupy 24.07 L volume
So 2666.67 g of Oxygen will occupy 24.07 X 2666.67 / 32
= 2005.84 L volume
So volume of Oxygen require will be 2005.84 L for complete burning of 1 kg of Coal having 100 % carbon.

  • -71

ACC. to Reacn. C + O2 -> CO2

1 mole O2 reacts with one mole C to give 1 moe CO2.therefore to react with 1 kg co2 moles of O2 required = 1000gm/12gm=83.33 moles

now, 

PV=nRT

​at NTP P=1bar R=0.082 and T=298K

therefore 1V=83.33x0.082x298

V=2002.16 ltres

  • -65
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