8.00g of CH4 is allowed to burn in the presence of 16.00g oxygen. How much (in g

Question

8.00g of CH4 is allowed to burn in the presence of 16.00g oxygen. How much (in grams) CH4, O2, CO2, and H2O remain after the reaction is complete?

Explanation / Answer

This is a combustion reaction (the products are CO2 and H2O) The balanced equation for this reaction is: CH4 + 2O2 -> CO2 + 2H2O Find the moles of each reactant to determine the limiting reagent. 8g CH4 / (16.05 g/mol) = 0.498 mol CH4 16 g O2 / (32 g/mol) = 0.500 mol O2 2 mol O2 are needed for 1 mol CH4 therefore, O2 is the limiting reagent, and once the reaction is complete, (0.498 - (.500/2)) = 0.248 mol CH4 will remain with no O2. 2 mol O2 yields 1 mol CO2 and 2 mol H20, so 0.250 mol CO2 are produced with 0.500 mol H20 multiply by each molar mass to determine the grams of each substance that remain after the reaction: 0.248 mol CH4 (16.05 g/mol) = 3.98 g 0.250 mol CO2 (44.01g/mol) = 11.0 g 0.500 mol H2O (18.02g/mol) = 9.01 g

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