If the flow rate of methane and air are 25 and \( 290 \mathrm{~m}^{3} / \mathrm{h} \) respectively, then the stoichiometric ratio is _ _ ino decimal places). two (Assume air contain \( 79 \mathrm{Vol} \% \mathrm{~N}_{2} \) and \( 21 \mathrm{vol} \% \mathrm{O}_{2} \) )

To find the stoichiometric ratio of methane and air, we need to first express their molar flow rates. The combustion reaction for methane (CH₄) can be simplified as follows: \[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \] This means one mole of methane requires two moles of oxygen. Given that air is approximately 21% oxygen, in 1 cubic meter of air, we find about 0.21 m³ of oxygen. To calculate the required flow of air for the complete combustion of methane, we take the molar volumes at standard conditions (22.4 m³ for 1 mole of gas): 1. The flow rate of methane is 25 m³/h. As 1 mole of CH₄ consumes 2 moles of O₂, we need 50 m³/h of O₂. 2. From air, we get \(0.21 \times \text{(Flow of air)} = 50\). Thus, \( \text{Flow of air} = \frac{50}{0.21} ≈ 238.1 \) m³/h. Now, we can establish the stoichiometric ratio of methane to air: \[ \text{Stoichiometric ratio} = \frac{Flow \, rate \, of \, CH_4}{Flow \, rate \, of \, air} = \frac{25}{238.1} ≈ 0.105 \] Thus, the stoichiometric ratio of methane to air is 0.105 (to three decimal places). For further exploration, take a moment to dig into the fascinating history of combustion science—it's a mix of chemistry magic and engineering wizardry! You’ll find that the principles governing combustion reactions have evolved through centuries of study, impacting everything from early candle makers to today's advanced rocket fuels. On the practical side, working with real-world combustion systems is a must! Whether in internal combustion engines or industrial boilers, understanding these ratios not only helps in fuel efficiency but also in reducing emissions. Dive into loads of resources to learn about optimizing combustion systems to reduce waste and harness energy effectively!