A gaseous fuel contains the following by Mass: 5% CO2; 40% H2; 40% CH4; 15% N2.

Question

A gaseous fuel contains the following by Mass: 5% CO2; 40% H2; 40% CH4; 15% N2. Determine the gravitational stoichiometric air and the % content of each WET product by mass.

Explanation / Answer

Let the total weight of the fuel be 100 grams. 5 grams CO2 40 grams H2 40 grams CH4 15 grams N2 Combustion equations of H2 and CH4 H2 + 0.5 ( O2+3.76N2) --------> H20 + 1.88 N2 (2g + 68.64g ---> 18g + 52.64 g) ( mass of each reactant and products, this implies 2 grams of h2 requires the respective mass of air and will give the mass of the products) CH4 + 2 ( O2+3.76N2) --------> CO2 + 2 H20 + 7.56 N2 (16g + 278.68 g ---> 44g + 36g + 211.68 g) ( mass of each reactant and products, this implies 16 grams of ch4 requires the respective mass of air and will give the mass of the products) Reaction 1: combustion of H2 2 grams of h2 will give 18 grams of h2o we have 40 grams of h2, which will give 40 *18/2 = 360 grams of h20. Reaction 2: Combustion of CH4 16 grams of ch4 will give 44 grams of co2 we have 40 grams of ch4, which will give 40 *44/16 = 110 grams of co2. 16 grams of ch4 will give 36 grams of h2o we have 40 grams of ch4, which will give 40 *36/16 = 90 grams of h2o. since we are looking for WET products, we don't consider water vapor in the final mixture. our final mixture will contain 5 grams + 110 grams (from combustion of CH4) = 115 grams of co2 15 grams + 52.64 grams (from combustion of H2) + 211.68 grams (combustion of CH4) = 279.32 of N2 Total mass of mixture = 115 + 279.32 = 394.32 grams 115 / 394.32 = 29.16 % by mass of CO2 279.32 / 394.32 = 70.84 % by mass of N2 29.16% CO2 and 70.84% N

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