Import Question JSON

$\text{Hint: Equation based question}$\n\n$\text{Step 1: Write the balanced chemical equation for the combustion of propane}$\n\n$\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$\n\n$\text{From the balanced equation, we can see that 5 moles of } \text{O}_2 \text{ react with 1 mole of } \text{C}_3\text{H}_8 \text{.}$\n\n$\text{Step 2: Apply Avogadro's Law and Gay-Lussac's Law}$\n\n$\text{According to Avogadro's Law, equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.}$\n\n$\text{By Gay-Lussac's Law, the ratio of volumes of gases in a reaction equals the ratio of their coefficients in the balanced equation when measured at the same temperature and pressure.}$\n\n$\text{Therefore, the volume ratio of } \text{C}_3\text{H}_8 \text{ to } \text{O}_2 \text{ is 1:5.}$\n\n$\text{Step 3: Calculate the volume of oxygen needed}$\n\n$\text{Given that we have 1 L of propane gas at } 0^{\circ}\text{C and 1 atm, the volume of oxygen needed will be:}$\n\n$\text{Volume of } \text{O}_2 = 1 \text{ L} \times 5 = 5 \text{ L}$\n\n$\text{Alternative method using mole concept:}$\n\n$\text{At STP (} 0^{\circ}\text{C and 1 atm), 1 mole of any gas occupies 22.4 L}$\n\n$\text{Moles of propane} = \frac{\text{Volume of propane}}{\text{Molar volume}} = \frac{1 \text{ L}}{22.4 \text{ L/mol}} = 0.0446 \text{ mol}$\n\n$\text{From the balanced equation, moles of oxygen needed} = 5 \times \text{moles of propane} = 5 \times 0.0446 = 0.223 \text{ mol}$\n\n$\text{Volume of oxygen at STP} = \text{moles of oxygen} \times \text{molar volume} = 0.223 \text{ mol} \times 22.4 \text{ L/mol} = 5 \text{ L}$\n\n$\text{Therefore, 5 L of oxygen gas is needed to completely burn 1 L of propane gas.}$