The general combustion reaction of alkanes is:
\[\text{C}_n\text{H}_{2n+2} + \frac{3n+1}{2}\text{O}_2 \rightarrow n\text{CO}_2 + (n+1)\text{H}_2\text{O}.\]
For methane (\(\text{CH}_4\)), the specific balanced equation is:
\[\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}.\]
Step 1: Molar mass of \(\text{CO}_2\)
The molar mass of \(\text{CO}_2\) is:
\[\text{Molar mass of } \text{CO}_2 = 12 + (2 \times 16) = 44 \, \text{g/mol}.\]
Step 2: Calculate moles of \(\text{CO}_2\)
The number of moles of \(\text{CO}_2\) in \(11 \, \text{g}\) is:
\[\text{Moles of } \text{CO}_2 = \frac{\text{Mass of } \text{CO}_2}{\text{Molar mass of } \text{CO}_2} = \frac{11}{44} = 0.25 \, \text{mol}.\]
Step 3: Relate \(\text{CH}_4\) to \(\text{CO}_2\)
From the balanced reaction:
\[1 \, \text{mole of } \text{CH}_4 \, \text{produces } 1 \, \text{mole of } \text{CO}_2.\]
Therefore, to produce \(0.25 \, \text{moles of } \text{CO}_2\), the moles of \(\text{CH}_4\) required are:
\[\text{Moles of } \text{CH}_4 = 0.25 \, \text{mol}.\]Step 4: Mass of \(\text{CH}_4\)
The mass of \(0.25 \, \text{moles of } \text{CH}_4\) can be verified as:
\[\text{Mass of } \text{CH}_4 = \text{Moles of } \text{CH}_4 \times \text{Molar mass of } \text{CH}_4 = 0.25 \times 16 = 4 \, \text{g}.\]
Conclusion:
The number of moles of methane required to produce \(11 \, \text{g of } \text{CO}_2\) is \(0.25 \, \text{mol}\).
Final Answer: (2).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)