What exact volume of oxygen gas at NTP is required for the complete combustion (oxidation) of of met — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

The balanced chemical equation for the combustion of methane is $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$. According to Avogadro's law, the stoichiometric coefficients represent the ratio of reacting volumes under identical conditions. Thus, $1\text{ volume}$ of $CH_4$ requires $2\text{ volumes}$ of $O_2$. Therefore, $2.0\text{ Litres}$ of $CH_4$ will strictly require $2.0 \times 2 = 4.0\text{ Litres}$ of $O_2$.