Question

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

• A. $22.400$ L
• B. $22.4\times {10}^3$L
• C. $1.86\times {10}^3$L
• D. $1000$ L

Answer 1

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

$\begin{array}{c}\text{At NTP conditions,}1\text{mole of a gas}\\ \text{occupies}22.4L\text{of volume. So, for}\\ \text{combustion of}1kg\text{of coal}\end{array}$

$\begin{array}{c}\text{number of mole of}coal=\frac{1000}{12}=83.333moles\\ \text{we will require}83.333moles\text{of oxygen}\left(O_2\right)\\ \text{which is occupies}=83.333\times 22.4\end{array}$

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