1)
дано
m(C3H6)=45 g
---------------------------
n(C3H7OH)-?
X 45
C3H7OH-->C3H6+H2O M(C3H7OH)=60 g/mol M(C3H6)=42 g/mol
60 42
X/60 = 45 /42
X=64.3 g
n(C3H7OH)=m/M = 64.3 / 60 = 1.07 mol
ответ 1.07 моль
2)
дано
n(Ag)=1.5 mol
-----------------------
m(CH3COH)-?
Xmol 1.5mol
CH3COH+Ag2O-->CH3COOH+2Ag
1 mol 2 mol
x/1 = 1.5/2
x=0.75 mol
m(CH3COH)=n(CH3COH)*M(CH3COH) , M(CH3COH)=44 g/mol
m(CH3COH)=0.75*44=33 g
ответ 33 г
3)
фруктоза - C6H12O6
M(C6H12O6)=180 g/mol
W(C)=12*6 / 180 *100%=40 %
W(H) = 1*12/180*100%=6.67%
W(O)= 16*6/180*100%=53.33%