Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
 \text{Kẽm + Axit Clohidric }\to\text{ Kẽm Clorua + khí Hidro} \\
b)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
c)\\
\text{Theo định luật bảo toàn khối lượng ta có :}\\
{m_{Zn}} + {m_{HCl}} = {m_{ZnC{l_2}}} + {m_{{H_2}}}\\
d)\\
\text{Ta có :}\\
{m_{Zn}} + {m_{HCl}} = {m_{ZnC{l_2}}} + {m_{{H_2}}}\\
 \Rightarrow {m_{HCl}} = 54,4 + 0,8 - 26 = 29,2g\\
e)\\
{N_{Zn}}:{N_{HCl}} = 1:2\\
{N_{Zn}}:{Z_{ZnC{l_2}}} = 1:1\\
{N_{HCl}}:{N_{{H_2}}} = 2:1\\
2)\\
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
b)\\
\text{Theo định luật bảo toàn khối lượng ta có :}\\
{m_{Al}} + {m_{{H_2}S{O_4}}} = {m_{A{l_2}{{(S{O_4})}_3}}} + {m_{{H_2}}}\\
c)\\
\text{Ta có :}\\
{m_{Al}} + {m_{{H_2}S{O_4}}} = {m_{A{l_2}{{(S{O_4})}_3}}} + {m_{{H_2}}}\\
 \Rightarrow {m_{{H_2}}} = {m_{Al}} + {m_{{H_2}S{O_4}}} - {m_{A{l_2}{{(S{O_4})}_3}}} = 5,4 + 29,4 - 34,2 = 0,6g\\
d)\\
{N_{Al}}:{N_{{H_2}S{O_4}}}:{N_{A{l_2}{{(S{O_4})}_3}}}:{N_{{H_2}}} = 2:3:1:3
\end{array}\)

 

1.

a. Kẽm + Axit clohidric → Kẽm clorua + khí Hidro.

b. `Zn+2HCl→ZnCl_2+H_2`

c. `m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}`

d. `m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}`

`⇒m_{HCl}=m_{ZnCl_2}+m_{H_2}-m_{Zn}`

`⇒m_{HCl}=54,4+0,8-26`

`⇒m_{HCl}=29,2(g)`

e. Tỉ lệ 3 cặp chất tùy chọn là:

`Zn:H_2=1:1`

`HCl:ZnCl_2=2:1`

`HCl:H_2=2:1`

2.

a. `2Al+3H_2SO_4→Al_2(SO_4)_3+3H_2`

b. `m_{Al}+m_{H_2SO_4}=m_{Al_2(SO_4)_3}+m_{H_2}`

c. `m_{Al}+m_{H_2SO_4}=m_{Al_2(SO_4)_3}+m_{H_2}`

`⇒m_{H_2}=m_{Al}+m_{H_2SO_4}-m_{Al_2(SO_4)_3}`

`⇒m_{H_2}=5,4+29,4-34,2`

`⇒m_{H_2}=0,6(g)`

d. Tỉ lệ giữa các chất:

`Al:H_2SO_4:Al_2(SO_4)_3:H_2=2:3:1:3`