`n_(H_2)=V/(24,79)=(7,437)/(24,79)=0,3(mol)`

`PTHH:`

`Zn+2HCl->ZnCl_2+H_2`

`0,3 ← 0,6 ←   0,3    ←   0,3(mol)`

`a)  m_(Zn)=n.M=0,3.65=19,5(g)`

`b)  m_(HCl)=n.M=0,6.36,5=21,9(g)`

`c)`

`@` Cách `1:`

`m_(ZnCl_2)=n.M=0,3.136=40,8(g)`

`@` Cách `2:` Áp dụng định luật BTKL, ta có:

`m_(Zn)+m_(HCl)=m_(ZnCl_2)+m_(H_2)`

`->  m_(ZnCl_2)=m_(Zn)+m_(HCl)-m_(H_2)=19,5+21,9-0,3.2=40,8(g)`

Đáp án + Giải thích các bước giải:

a.

Phản ứng xảy ra:

$Zn(s)+2HCl(aq) \to ZnCl_2(aq)+H_2(g)$

$n_{H_2}=\dfrac{7,437}{24,79}=0,3(mol)$

$n_{Zn}=n_{H_2}=0,3(mol)$

$m_{Zn}=0,3.65=19,5(g)$

b.

$n_{HCl}=2.n_{H_2}=2.0,3=0,6(mol)$

$m_{HCl}=0,6.36,5=21,9(g)$

c.

Cách 1:

$n_{ZnCl_2}=n_{H_2}=0,3(mol)$

$m_{ZnCl_2}=0,3.136=40,8(g)$

Cách 2:

$\xrightarrow[]{\texttt{BTKL}}:m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}$

$ \to 19,5+21,9=m_{ZnCl_2}=0,3.2$

$ \to m_{ZnCl_2}=40,8(g)$