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CnH2nOn + O2 ---> CO2 + H2O
0.5 mola CnH2nOn ------- 56dm3
1 mol CnH2nOn -------- x
x = 112dm3
112dm3 : 22,4dm3 = 5 [moli CO2]
liczba moli CO2 jest równa n, a więc ten cukier to C5H10O5 - pentoza np. ryboza