`#AkaShi`

Bài `5` :

`Fe_2O_4  + 4H_2SO_4 -> FeSO_4 + Fe_2(SO_4)_3 + 4H_2O`

`Fe_xO_y + 2yHCl -> xFeCl_{(2y)/(x)}  + yH_2`

`Mg(OH)_2 + 2HNO_3 -> Mg(NO_3)_2 + 2H_2O`

`CaCO_3 + 2HCl -> CaCl_2 + H_2O + CO_2↑`

`H_2SO_4 + 2KOH -> K_2SO_4 + 2H_2O`

`Ba(NO_3)_2 + H_2SO_4 -> BaSO_4↓ + 2HNO_3`

`AgNO_3 + HCl -> AgCl↓ + HNO_3`

`2M + 2xHCl -> 2MCl_x + nH_2↑`

Bài `6` :

`a)` PTHH:

`Mg + 2HCl -> MgCl_2 + H_2↑`

`→n_{H_2}=(4,48)/(22,4)=0,2 (mol)`

`→n_{Mg}=n_{H_2}=0,2 (mol)`

`→m_{Mg}=a (g)=0,2xx24=4,8 (g)`

`b)`

`→n_{HCl}=0,2xx2=0,4 (mol)`

`→m_{HCl}=0,4xx36,5=14,6 (g)`

`→C%_{HCl}=(14,6)/200xx100%=7,3%`

`c)`

`→m_{dd\ sau\ pư}=4,8+200-0,2xx2=204,4 (g)`

`→n_{MgCl_2}=n_{H_2}=0,2 (mol)`

`→C%_{MgCl_2}=(0,2xx95)/(204,4)xx100%=9,3%`

Em tham khảo nha:

\(\begin{array}{l}
5)\\
a)\,F{e_3}{O_4} + 4{H_2}S{O_4} \to FeS{O_4} + F{e_2}{(S{O_4})_3} + 4{H_2}O\\
b)\,F{e_x}{O_y} + 2yHCl \to xFeC{l_{\dfrac{{2y}}{x}}} + y{H_2}\\
c)\,Mg{(OH)_2} + 2HN{O_3} \to Mg{(N{O_3})_2} + 2{H_2}O\\
d)\,CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
e)\,2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
g)\,Ba{(N{O_3})_2} + {H_2}S{O_4} \to BaS{O_4} + 2HN{O_3}\\
h)\,AgN{O_3} + HCl \to AgCl + HN{O_3}\\
i)\,2M + 2nHCl \to 2MC{l_n} + n{H_2}\\
6)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,4\,mol\\
{C_\% }HCl = \dfrac{{0,4 \times 36,5}}{{200}} \times 100\%  = 7,3\% \\
c)\\
{n_{Mg}} = {n_{{H_2}}} = 0,2\,mol\\
{m_{{\rm{dd}}spu}} = 0,2 \times 24 + 200 - 0,2 \times 2 = 204,4g\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,2\,mol\\
{C_\% }MgC{l_2} = \dfrac{{0,2 \times 95}}{{204,4}} \times 100\%  = 9,3\% 
\end{array}\)