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Consider the reaction. CH4(g)+2O2(g) -> CO2(g)+2H20(g) How many grams of methane should be burned in an excess of oxygen at STP to obtain 5.6 L of carbon dioxide? A. 2.0 g B. 4.0 g C. 16.0 g D. 32.0 g
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molar masses methane = 16 g/mol carbon dioxide =44 g/mol One mole of any gas at S.T.P. occupies the same volume which is 22.4 L. divide this into 5.6 L to find how many grams we have: 5.6/22.4 = 0.25, so there are (0.25)(44 g) = 11 g. we see that 1 mole of CO2 is produced for 1 mole of CH4 Since we produce 11 g = 0.25 mol of CO2 we must start with 0.25 mol of CH4 (0.25)(16 g) = 4.0 g.
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