Mass of carbon = 1000g
Moles of carbon = 1000/12=83.34moles
C+O2 -> CO2
1moles C requires 1mole O2 to burn.
So, 83.34 moles C requires 83.34 moles O2.
At NTP, 1 mole gas occupies 22.4litres.
So, 83.34 moles occupies 83.34×22.4
= 1.86×10^3L

Calculation of the Volume of NTP Oxygen Required to React Completely Burn 1 kg of Coal (100% Carbon)

Assumption: The reaction is assumed to be at standard temperature and pressure (STP), which is 0°C and 1 atm, respectively.

Step 1: Write the Balanced Chemical Equation of the Reaction
The balanced chemical equation of the reaction between carbon and oxygen is:

C + O2 → CO2

This equation indicates that one mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide.

Step 2: Calculate the Molecular Weight of Carbon
The molecular weight of carbon is 12 g/mol.

Step 3: Calculate the Number of Moles of Carbon in 1 kg of Coal
Since coal is 100% carbon, the number of moles of carbon in 1 kg of coal is:

1 kg of coal × (1 mol of carbon/12 g of carbon) = 83.33 moles of carbon

Step 4: Calculate the Number of Moles of Oxygen Required for Complete Combustion of 1 kg of Coal
From the balanced chemical equation, the stoichiometric ratio of carbon to oxygen is 1:1. Therefore, the number of moles of oxygen required for complete combustion of 1 kg of coal is also 83.33 moles.

Step 5: Calculate the Volume of NTP Oxygen Required for Complete Combustion of 1 kg of Coal
The volume of NTP oxygen required for complete combustion of 1 kg of coal can be calculated using the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant (0.082 L·atm/mol·K), and T is the temperature.

At STP, the pressure is 1 atm and the temperature is 0°C or 273 K. Therefore, the volume of NTP oxygen required is:

V = nRT/P = (83.33 mol)(0.082 L·atm/mol·K)(273 K)/(1 atm) = 1866 L or 1.866 m3

Therefore, the volume of NTP oxygen required for complete combustion of 1 kg of coal is 1.866 m3.