Question

In order to get maximum calorific output, a burner should have an optimum fuel to oxygen ratio which corresponds to 3 times as much oxygen as is required theoretically for complete combustion of fuel. A burner which has been adjusted for methane as fuel (with $X$ litre/hour of $C{H}_4$ and $6X$ litre/hour of $O_2$) is to be readjusted for butane $C_4{H}_{10}$.

Assume that losses due to incomplete combustion, etc. are the same for both fuels and that gases behave ideally. Heat of combustion for $C{H}_4=809kJmo{l}^{-1}$ and $C_4{H}_{10}=2878kJmo{l}^{-1}$.

In order to get same calorific output, the rate of supply of oxygen would be:

• A. $5.48\left(X\right)litre/hr$
• B. $3.56\left(X\right)litre/hr$
• C. $6.59\left(X\right)litre/hr$
• D. None of these

Answer 1

Answer: (A)

$5.48\left(X\right)litre/hr$

$\text{Given}:$ Heat of combustion for $C{H}_4$ = 809kJ/mol, $C_4{H}_{10}=2878kJ/mol$

$\text{Solution}:$

In 1h, xL of $C{H}_4$ required 6x L of $O_2$

$\Delta H\left(combustion\right)$ of $C{H}_4=809kJmo{l}^{-1}$

$⟹\frac{809}{24.48}kJmo{l}^{-1}$ at 1.0 atm and ${25}^{o}C$

$\Delta H\left(combustion\right)$ of $C_4{H}_{1}0=2878kJmo{l}^{-1}$

$⟹\frac{2878}{24.48}kJ{L}^{-1}$ at 1.0 atm and ${25}^{o}C$

xL of $C{H}_4$ produces $\frac{809}{24.48}xkJ$

Now this much enegry has to be provided by burning of $C_4{H}_{1}0$.

$⟹\frac{809}{24.48}\left(x\right)kJ$ will be provided by

$\frac{809}{24.48}\left(x\right)\times \frac{24.48}{2878}-\left(0.28x\right)L$ of $C_4{H}_{1}0$

$C_4{H}_{1}0+\frac{13}{2}{O}_2\to 4C{O}_2+5H_{2}O$

$⟹1mol{C}_4{H}_{1}0=\frac{13}{2}mol{O}_2$

$⟹3\times \frac{13}{2}$ times of $O_2$ is required per mol. $⟹$ Rate of $O_2$ per hour $=\left(0.28x\right)\times \frac{39}{2}=\left(5.48x\right)L{O}_2$

$\text{Hence A is the correct option}$