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Question 3.92: Deal with the methane and propane separately, and combine th......

Deal with the methane and propane separately, and combine the results. Balanced equations are needed for each hydrocarbon. The total mass and the percentages will give the mass of each hydrocarbon . The mass of each hydrocarbon is changed to moles, and through the balanced chemical equation the amount of \text{CO}_2 produced by each gas may be found. Summing the amounts of \text{CO}_2 gives the total from the mixture .

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The balanced chemical equations are :
Methane : \quad\quad \text{CH}_4\text{(g) + 2 O}_2\text{(g)} → \text{CO}_2\text{(g) + 2 H}_2\text{O}(l)
Propane : \quad\quad \text{C}_3\text{H}_8\text{(g) + 5 O}_2\text{(g)} → 3\text{ CO}_2\text{(g) + 4 H}_2\text{O}(l)
Mass of \text{CO}_2 from each :

\text{Methane: }(200.\text{ g Mixture})\left\lgroup\frac{25.0\%}{100\%} \right\rgroup \left\lgroup\frac{1\text{ mol CH}_4}{16.04\text{ g CH}_4} \right\rgroup \left\lgroup\frac{1\text{ mol CO}_2}{1\text{ mol CH}_4} \right\rgroup \left\lgroup\frac{44.01\text{ g CO}_2}{1\text{ mol CO}_2} \right\rgroup =137.188\text{ g CO}_2 \\ \text{Propane: }(200.\text{ g Mixture})\left\lgroup\frac{75.0\%}{100\%} \right\rgroup \left\lgroup\frac{1\text{ mol C}_3\text{H}_8}{44.09\text{ g C}_3\text{H}_8} \right\rgroup \left\lgroup\frac{3\text{ mol CO}_2}{1\text{ mol C}_3\text{H}_8} \right\rgroup \left\lgroup\frac{44.01\text{ g CO}_2}{1\text{ mol CO}_2} \right\rgroup =449.183\text{ g CO}_2 \\ \text{Total CO}_2=137.188\text{ g + 449.183 g = 586.318 = 586 g CO}_2

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