CH3-CH2-COOH+HO-CH3= CH3-CH2-CO-O-CH3+H2O
дано
V(H2)-100 л
m(1лH2)-0.09r
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Q-?
решение:
2H2+O2=2H2O+572кДж 4г
m(100л H2)=0.09 г •100=9 г
Ba(OH)2+2HNO3=Ba(NO3)2+2H2O
1. D(j)(O2)= 2.75 ==> M(j)= 2.75 * 32 = 88 г/моль
CnH2n+1OH = 88
14n + 18 = 88
14n = 70
n = 5
C5H13OH — пентан-1-ол.
2. D(j)(N2)= 2.61 ==> M(j)= 2.61 * 28 = 73 г/моль
NH2 + R = 73
16 + R = 73
R = 57
n(C) = 57/12 = 4 ( ост. 9 ) ==> R = C4H9
C4H9NH2 — бутанамин.
3. M(C5H10O4)= 134 г/моль
W(C) = 60/134 = 44.78%
W(H) = 10/134 = 7.47%
W(O) = 100% - 44.78% - 7.47% = 47.75%