Kerala KEAM 5 June 2024 Solved Paper

The combustion of methane (CH₄) in excess oxygen (O₂) forms carbon dioxide (CO₂) and water (H₂O). The balanced chemical equation for the combustion reaction is:$$CH_4 (g) + 2 O_2 (g) \rightarrow CO_2 (g) + 2 H_2O (g)$$From the equation, we can see that 1 mole of methane produces 2 moles of water.Step 1: Calculate the molar mass of methane (CH₄) and water (H₂O):- Molar mass of CH₄ = $12 + (4 \times 1) = 16 \, g/mol$- Molar mass of H₂O = $(2 \times 1) + 16 = 18 \, g/mol$Step 2: Convert the mass of methane to moles:We are given 1.6 g of methane. To calculate the number of moles of methane: $$Moles of CH₄ = \frac{1.6 \, g}{16 \, g/mol} = 0.1 \, mol$$Step 3: Use the mole ratio from the balanced equation to find the moles of water:From the balanced equation, 1 mole of methane produces 2 moles of water. Therefore, 0.1 mole of methane will produce: $$Moles of H₂O = 0.1 \, mol \times 2 = 0.2 \, mol$$Step 4: Calculate the mass of water:Now, to find the mass of water, multiply the moles of water by its molar mass: $$Mass of H₂O = 0.2 \, mol \times 18 \, g/mol = 3.6 \, g$$Thus, the correct answer is option (D), 3.6 g. Quick Tip: Always use the mole ratio from the balanced equation to relate the moles of reactants and products. Then use the molar masses to convert between grams and moles.