Đáp án:

 Bài 3: a/3,37g

b/1,62

Bài 4:  a/

$\%m_{Mg}=41,86\%⇒\%m_{CaCO_3}=58,14\ \%$

b/ $V_{HCl}= 2\ lít$

Giải thích các bước giải:

 Bài 3: $n_{H_2}=\dfrac{0,04}{2}=0,02\ mol; n_{HCl}=0,052\ mol$

$Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\\x\hspace{2cm}6x\\Zn+2HCl\to ZnCl_2+H_2\\0,02\hspace{0,5cm}0,04\hspace{3cm}0,02$

$⇒6x = 0,52 - 0,04 ⇒ x = 0,002\ mol$

a/ $m_{muối}=m_{FeCl_3}+m_{ZnCl_2}=0,002×2×162,5 +0,02×136=3,37g$

b/ $x = m_{Fe_2O_3}+m_{Zn}=0,002×160+0,02×65=1,62$

Bài 4: $n_{Khí}=\dfrac{4,48}{22,4}=0,02\ mol$

$Mg+2HCl\to MgCl_2+H_2\\x\hspace{5cm}x\\CaCO_3+2HCl\to CaCl_2+CO_2+H_2O\\y\hspace{6cm}y$

Theo bài ra, ta có hệ: $\left \{ {{24x+100y=8,6} \atop {x+y=0,2}} \right.$ ⇔$\left \{ {{x=0,15} \atop {y=0,05}} \right.$ 

a/$m_{Mg}=0,15×24=3,6\%m_{Mg}=\dfrac{3,6}{8,6}×100=41,86\%\\⇒\%m_{CaCO_3}=58,14\ \%$

b/ $n_{HCl}=2×n_{Mg}+2×n_{CaCO_3}=0,4⇒V_{HCl}=\dfrac{0,4}{0,2}= 2\ lít$