Volume at NTP of oxygen required to completely burn 1 kg of coal (100% carbon):

22.400

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22.4 \times 10^{3}

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1.86 \times 10^{3}

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1000

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Solution

The correct option is C $1.86 \times 10^{3}$ L
$\begin{matrix} At NTP conditions, 1 mole of a gas occupies 22.4 L of volume. So, for combustion of 1 k g of coal \end{matrix}$

$\begin{matrix} number of mole of c o a l = \frac{1000}{12} = 83.333 m o l e s we will require 83.333 m o l e s of oxygen (O_{2}) which is occupies = 83.333 \times 22.4 \end{matrix}$

$= 1.866 \times 10^{3} L of volume of NTP.$

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Q. Calculate the volume of oxygen at NTP that would be required to convert 5.2 L of carbon monoxide to carbon dioxide.

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Volume at NTP of oxygen required to completely burn 1 kg of coal (100% carbon):

The correct option is C 1.86×103L At NTP conditions, 1 mole of a gas occupies 22.4 L of volume. So, for combustion of 1 kg of coal number of mole of coal=100012=83.333moles we will require 83.333moles of oxygen (O2) which is occupies =83.333×22.4 =1.866×103 L of volume of NTP.