Đáp án:

$\rm a)\\n_{H_2}=\dfrac{2,24}{22,4}=0,1\ (mol)$

Phương trình phản ứng:

$\rm Zn+H_2SO_4\to ZnSO_4+H_2$

Theo phương trình:

$\rm n_{H_2SO_4}=n_{H_2}=0,1\ (mol)$

$\rm\to m_{dd\ H_2SO_4}=\dfrac{0,1×98}{19,6\%}=50\ (g)\\b)$

Theo phương trình:

$\rm n_{Zn}=n_{H_2}=0,1\ (mol)$

$\rm\to\%m_{Zn}=\dfrac{0,1×65}{10,5}×100\%=61,9\%\\\to\%m_{Cu}=100-61,9=38,1\%\\c)$

Theo phương trình:

$\rm n_{ZnSO_4}=n_{H_2}=0,1\ (mol)$

$\rm\to m_{ZnSO_4}=0,1×161=16,1\ (g)$

$PTHH : Zn +H_2SO_4 \xrightarrow{} ZnSO_4 +H_2$

$n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$

$=> n_{H_2SO_4}=0,1(mol)$

$⇒ m_{H_2SO_4}=\dfrac{0,1×98}{19,6}×100 =50 (g)$

b) $n_{Zn}=n_{H_2}=0,1(mol)$

$⇒m_{Zn}=0,1 ×65=6,5(g)$

%$m_{Zn}=\dfrac{6,5}{10,5}×100≈61,9 $%

⇒%$m_{Cu}=100-61,9=38,1$%

c)$n_{ZnSO_4}=n_{H_2}=0,1(mol)$

$⇒m_{ZnSO_4}=0,1×161=16,1(g)$