Bài giải :

Bài 19 :

Sửa: `1,79(l)CO_2` thành `1,792(l)CO_2`

a.

`-n_{N_2}=\frac{V_{N_2}}{22,4}=\frac{5,6}{22,4}=0,25(mol)`

`-n_{H_2}=\frac{V_{H_2}}{22,4}=\frac{4,48}{22,4}=0,2(mol)`

`-n_{CO_2}=\frac{V_{CO_2}}{22,4}=\frac{1,792}{22,4}=0,08(mol)`

b.

`-m_{N_2}=n_{N_2}.M_{N_2}=0,25.28=7(g)`

`-m_{H_2}=n_{H_2}.M_{H_2}=0,2.2=0,4(g)`

`-m_{CO_2}=n_{CO_2}.M_{CO_2}=0,08.44=3,52(g)`

`⇒m_{hh..A}=m_{N_2}+m_{H_2}+m_{CO_2}`

                  `=7+0,4+3,52=10,92(g)`

Bài 20 :

a.

`-m_{Al}=n_{Al}.M_{Al}=0,9.27=24,3(g)`

Vì $d_{Al}=2,7(g/cm^3)$

`⇒V_{Al}=\frac{m_{Al}}{d}=\frac{24,3}{2,7}=9(cm^{3})`

b.

`-m_{Cl_2}=n_{Cl_2}.M_{Cl_2}=1,25.71=88,75(g)`

`-V_{Cl_2}(đktc)=n_{Cl_2}.22,4=1,25.22,4=28(l)`

c.

`-m_{NH_3}=n_{NH_3}.M_{NH_3}=0,86.17=14,62(g)`

 Bài 21 :

a.

`-n_{N_2}=\frac{m_{N_2}}{M_{N_2}}=\frac{28}{28}=1(mol)`

`-n_{NO}=\frac{m_{NO}}{M_{NO}}=\frac{15}{30}=0,5(mol)`

`⇒n_{hh}=n_{N_2}+n_{NO}=1+0,5=1,5(mol)`

`⇒V_{hh}(đktc)=1,5.22,4=33,6(l)`

b.

Vì `V_{H_2O}=0,8(l)=800(ml)`

Mà $d_{H_2O}=1(g/cm^{3})$

`⇒m_{H_2O}=V_{H_2O}.d_{H_2O}=800.1=800(g)`

`⇒n_{H_2O}=\frac{m_{H_2O}}{M_{H_2O}}=\frac{800}{18}=\frac{400}{9}(mol)`