дано

m(CxHyOz) =2.3 g

m(CO2) = 4.4 g

m(H2O) = 2.7 g

D(возд) = 1.59

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CxHyOz-?

M(CxHy) = D(возд)*29= 46 g/mol

n(C) = n(CO2) = m/M = 4.4 / 44 = 0.1 mol

m(C) = n(C)*M(C) = 0.1*12 =1.2 g

n(H) = 2n(H2O) = 2*( 2.7 / 18) = 0.3 mol

m(H) = n(H)*M(H) = 0.3*1 = 0.3 g

m(O) = m(CxHyOz) - (m(C)+mO)) = 2.3 - 1.2-0.3 = 0.8 g

M(O) = 16 g/mol

n(O) = m/M = 0.8 / 16 = 0.05 g/mol

n(C) : n(H) : n(O) = 0.1 :0.3 : 0.05 = 2 : 6 :1

C2H5OH

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