Alexei78
1) дано m(C3H7OH)=160 g m(CH3COOH)=120 g η(эфир)=95% ---------------------- m(эфир)-? M(C3H7OH)=60 g/mol M(CH3COOH)=60 g/mol n(C3H7OH)=m/M=160/60=2.67 mol n(CH3COOH)=m/M=120/60 = 2 mol n(C3H7OH)>n(CH3COOH) 120 X CH3COOH+C3H7OH-->CH3COOC3H7+H2O M(CH3COOC3H7)=102 g/mol 60 102 120/60=X/102 X=204 g m(эфир)=204*95%/100%=193.8 g ответ 193.8 г
2) дано m(грязн Mg)=80 g W(прим)=15% CH3COOH ------------------ V(H2)-? m(чист Mg)=80-(80*15%/100%)=68 g 68 X 2Mg+2CH3COOH-->(CH3COO)2Mg+H2 M(Mg)=24 g/mol , Vm=22.4L/mol 2*24 22.4 68/48 = x/22.4 X=31.73 L ответ 31.73л 3) дано m(ppaC2H5-COOH)=200 g W=30% ZnO --------------- m((C2H5COO)2Zn)-? m(C2H5COOH)=200*30%/100%=60 g 60 X 2C2H5COOH+ZnO-->(C2H5COO)2Zn+H2O 2*74 211 M(C2H5COOH)=74 g/mol M((C2H5COO)2Zn)=211 g/mol 60/148 = X/211 X=85.54 g ответ 85.54 г
Answers & Comments
дано
m(C3H7OH)=160 g
m(CH3COOH)=120 g
η(эфир)=95%
----------------------
m(эфир)-?
M(C3H7OH)=60 g/mol
M(CH3COOH)=60 g/mol
n(C3H7OH)=m/M=160/60=2.67 mol
n(CH3COOH)=m/M=120/60 = 2 mol
n(C3H7OH)>n(CH3COOH)
120 X
CH3COOH+C3H7OH-->CH3COOC3H7+H2O M(CH3COOC3H7)=102 g/mol
60 102
120/60=X/102
X=204 g
m(эфир)=204*95%/100%=193.8 g
ответ 193.8 г
2)
дано
m(грязн Mg)=80 g
W(прим)=15%
CH3COOH
------------------
V(H2)-?
m(чист Mg)=80-(80*15%/100%)=68 g
68 X
2Mg+2CH3COOH-->(CH3COO)2Mg+H2 M(Mg)=24 g/mol , Vm=22.4L/mol
2*24 22.4
68/48 = x/22.4
X=31.73 L
ответ 31.73л
3)
дано
m(ppaC2H5-COOH)=200 g
W=30%
ZnO
---------------
m((C2H5COO)2Zn)-?
m(C2H5COOH)=200*30%/100%=60 g
60 X
2C2H5COOH+ZnO-->(C2H5COO)2Zn+H2O
2*74 211
M(C2H5COOH)=74 g/mol
M((C2H5COO)2Zn)=211 g/mol
60/148 = X/211
X=85.54 g
ответ 85.54 г