(2,0 Đ) C2H5OH + NA  C2H5ONA + ½ H20,25N H2= 0,03 MOL => NC2...

Câu 6: (2,0 đ)

C

2

H

5

OH + Na  C

2

H

5

ONa + ½ H

2

0,25

n H

2

= 0,03 mol => nC

2

H

5

OH = 0,06 mol

=>mC

2

H

5

OH = 0,06 x 46 =2,76 gam

n Ag= 0,1 mol => n HCHO = 0,05

 m HCHO = 0,05 x 30 = 1,5 gam

 m hh = 2,76 + 1,5 =4,26 gam

 % (m) C

2

H

5

OH = 64,8%

 %(m) HCHO = 35,2%

C

2

H

5

OH + CuO  CH

3

CHO + Cu + H

2

O

m CuO =0,06 x 80 = 4,8 gam