$a)$

$PTHH$: $Zn$ $+$ $H_{2}SO_{4}$ $→$ $ZnSO_{4}$ $+$ $H_{2}$

$b)$

$n_{H_{2}}$ $=$$\frac{4,48}{22,4}$ $=$ $0,2(mol)$

$n_{Zn}$ $=$ $n_{H_{2}}$ $=$ $0,2(mol)$

$m_{Zn}$ $=$ $0,2$.$65$ $=$$13(g)$

$m_{Cu}$ $=$ $20-13$ $=$ $7(g)$

$c)$

$n_{H_{2}SO_{4}}$ $=$ $n_{H_{2}}$ $=$ $0,2(mol)$

$m_{H_{2}SO_{4}}$ $=$ $0,2$.$98$ $=$ $19,6(g/mol)$

$C$%$_{H_{2}SO_{4}}$ $=$ $\frac{19,6}{196}$ . $100$% $=$ $10$%

Đáp án:

Giải thích các bước giải:

$a)$ Phương trình hóa học:

$Zn+H_2SO_4\to ZnSO_4+H_2$

$b)$ $n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)$

Theo phương trình:

$n_{Zn}=n_{H_2}=0,2(mol)$

$m_{Zn}=0,2.65=13(g)$

$m_{Cu}=20-13=7(g)$

$c)$

Theo phương trình:

$n_{H_2SO_4}=n_{H_2}=0,2(mol)$

$C\%_{H_2SO_4}=\dfrac{0,2.98}{196}.100\%=10\%$