Bài giải :

Bài 4 :

`a.`

`-n_{Fe}=\frac{m_{Fe}}{M_{Fe}}=\frac{1,2}{56}=0,2(mol)`

Phương trình hóa học :

 `Fe+2HCl→FeCl_2+H_2↑`

`0,2` `→` `0,4`       `0,2`          `0,2`          `(mol)`

`⇒V_{H_2}(đktc)=n_{H_2}.22,4=0,2.22,54=4,48(l)`

`b.`

`-m_{HCl}=n_{HCl}.M_{HCl}=0,4.36,5=14,6(g)`

`c.`

`m_{FeCl_2}=n_{FeCl_2}.M_{FeCl_2}=0,2.127=25,4`(g)`

Bài 5 :

`a`.

`-n_{Al}=\frac{2,4.10^{23}}{6.10^{23}}=0,04(mol)`

 $4Al+3O_2\xrightarrow{t^o}2Al_2O_3 $

`0,04→0,03`   `0,02`       `(mol)`

`⇒V_{O_2}(đktc)=n_{O_2}.22,4=0,03.22,4=0,672(l)`

`⇒V_{kk}=5.V_{O_2}=5.0,672=3,36(l)`

`b.`

`-m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=0,02.102=2,04(g)`

Em tham khảo nha:

\(\begin{array}{l}
4)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{11,2}}{{56}} = 0,2mol\\
{n_{{H_2}}} = {n_{Fe}} = 0,2\,mol\\
{V_{{H_2}}} = n \times 22,4 = 0,2 \times 22,4 = 4,48l\\
b)\\
{n_{HCl}} = 2{n_{Fe}} = 0,4\,mol\\
{m_{HCl}} = n \times M = 0,4 \times 36,5 = 14,6g\\
c)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,2\,mol\\
{m_{FeC{l_2}}} = n \times M = 0,2 \times 127 = 25,4g\\
5)\\
a)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
{n_{Al}} = \dfrac{N}{{6 \times {{10}^{23}}}} = \dfrac{{2,4 \times {{10}^{22}}}}{{6 \times {{10}^{23}}}} = 0,04\,mol\\
{n_{{O_2}}} = \dfrac{{0,04 \times 3}}{4} = 0,03\,mol\\
{V_{{O_2}}} = n \times 22,4 = 0,03 \times 22,4 = 0,672l\\
{V_{kk}} = {V_{{O_2}}} \times 5 = 0,672 \times 5 = 3,36l\\
b)\\
{n_{A{l_2}{O_3}}} = \dfrac{{{n_{Al}}}}{2} = \dfrac{{0,04}}{2} = 0,02\,mol\\
{m_{A{l_2}{O_3}}} = n \times M = 0,02 \times 102 = 2,04g
\end{array}\)