Question

Volume at NTP of oxygen required to completely burn 1 kg of coal

Answer 1

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

Answer 2

Answer: Volume at NTP of oxygen required to completely burn 1 kg of coal  is 1866 L.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}$

$\text{Number of moles of carbon}=\frac{1000g}{12g/mol}=83.3moles$

According to the balance equation for combustion of coal:

$C+O_2\rightarrow CO_2$

It is seen that 1 mole of coal requires 1 mole of Oxygen gas for combustion

Thus 83.3 moles of coal will require=$\farc{1}{1}\times 83.3=83.3 moles$ of oxygen

Now 1 mole of every gas occupies 22.4 L at NTP

83.3 moles of $O_2$ occupies=$\frac{22.4}{1}\times 83.3=1866L$