дано
m(CxHyOz) = 11.5 g
V(CO2) = 5.6 L
m(H2O) = 4.5 g
D(возд) = 1.59
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CxHyOz-?
M(CxHyOz) = D()* 29 = 1.59 * 29 = 46.11 g/mol
n(C) = n(CO2) = V/Vm = 5.6 / 22.4 = 0.25 mol
n(H) = 2*n(H2O) = 2*(m(H2O) / M(H2O)) = 2*( 4.5 / 18) = 0.5 mol
M(C) = 12 g/mol
m(C) = n(C)*M(C) = 0.25 * 12 = 3 g
M(H) = 1 g/mol
m(H) = n(H)*M(H) = 0.5 * 1 = 0.5 g
m(O) = m(CxHyOz) - (m(C)+m(H)) = 11.5 - (3+0.5) = 8 g
M(O) = 16 g/mol
n(O) = m(O) / M(O) = 8 / 16 = 0.5 mol
n(C) : n(H) : n(O) = 0.25 : 0.5 : 0.5 = 1:2:2
CH2O2 ==> HCOOH
ответ : Муравьиная кислота