Прошу, помогите!!! Очень безотлагательно надобно!!!1.

1.
дано
m(CxHyOz) = 1.8 g
m(CO2) = 3.96 g
m(H2O) = 2.16 g
D(H2) = 30 
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M(CxHyOz)-?

M(CxHyOz) = D(H2) * 2 = 30*2 = 60 g/mol
M(CO2) = 44 g/mol
n(CO2) = m/M = 3.96 / 44 = 0.09 mol
n(C) = n(CO2) = 0.09 mol
n(C) = 0.09 mol
M(C)= 12 g/mol
m(C) = n*M= 0.09*12 = 1.08 g
M(H2O) = 18 g/mol
n(H2O) =m/M =  2.16/18 = 0.12 mol
n(H2) = 2n(H2O)  , n(H2) = 2*0.12 = 0.24 mol
M(H)= 1 g/mol
m(H) = n*M = 0.24 * 1 = 0.24 g
m(O) = m(CxHyOz)  - (m(C)+m(H)) = 1.8-(1.08 + 0.24) = 0.48 g
M(O) = 16 g/mol
n(O) = m/M= 0.48 / 16 = 0.03 mol
CxHyOz = 0.09 : 0.24 : 0.03 = 3:8:1
C3H8O
C3H7OH  M(C3H7OH) = 12*3+1*7+16+1) = 60 g/mol
ответ ПРОПАНОЛ

2)
дано
W(C) = 85.7% = 0.857
W(H) = 14.3% = 0.143
D(H2) = 28
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M(CxHy) -?

M(CxHy) = D(H2) *28 = 2*28 = 56 g/mol

W(CxHy) / M(CxHy) = W(C) / Ar(C) : W(H)/ Ar(H)
CxHy = 4  : 8
C4H8
ответ C4H8

3)
дано
m(CxHy) = 2.34 g
V(CO2) = 4.032 L
D(H2) = 39 
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M(C6H6)-?
M(CxHy) = D(H2)*39 = 2*39 = 78 g/mol
n(CO2) =V(CO2) / Vm = 4.032 / 22.4 = 0.18 mol
n(C) = n(CO2) = 0.18 mol
M(C) = 12 g/mol
m(C) = n*M = 0.18 * 12 = 2.16 g
m(H) = m(CxHy) - m(C) = 2.34 - 2.16 = 0.18 g
M(H) = 1 g/mol
n(H) = m/M = 0.18 / 1 = 0.18 mol
CxHy =n(C) : n(H) = 0.18 : 0.18  = 1:1
CH
M(CH) = 13 g/mol
n(CH) = M(CxHy) / M(CH) = 78 / 13 = 6 раз
M(CxHy) = 6*M(CH)= 6*13 = 78
CxHy = C6H6
ответ БЕНЗОЛ