Đáp án:
Câu 3:
1. $Fe_3O_4+4CO\to 3Fe+4CO_2\qquad(1)$
$ZnO+CO\to Zn+CO_2\qquad(2)$
$CO_2+Ca(OH)_2→CaCO_3(↓)+H_2O\qquad(3)$
$n_{CaCO_3}=\dfrac{60}{100}=0,6(mol)$
Theo PTHH $(1)→n_{CO_2}=n_{CaCO_3}=0,6(mol)$
$\to n_{CO}=n_{CO_2}=0,6(mol)$
$\to V=0,6.22,4=13,44(l)$
Gọi $x;y$ là số mol $Fe_3O_4; ZnO$
Ta có hệ phương trình
$232x+81y=38,4$
$4x+y=0,6$
Giải được: $x=0,1;y=0,2$
$→m_{Fe_3O_4}=232.0,1=23,2(g)$
$→m_{ZnO}=39,4-23,2=16,2(g)$
2. $n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)$
$a)$
\[MgO + 2HCl\to MgCl_2+H_2O\qquad(1)\]
\[Mg+2HCl\to MgCl_2+H_2\qquad(2)\]
\[MgCl_2+2NaOH\to Mg(OH)_2+2NaCl\qquad(3)\]
\[Mg(OH)_2\to MgO+H_2O\qquad(4)\]
$b)$
Theo PTHH $(1)\to n_{Mg}=n_{H_2}=0,2(mol)$
\[\to m_{Mg}=0,2\times 24=4,8(g)\]
\[\to m_{MgO}=8,4-4,8=3,6(g)\]
\[\to n_{MgO}=\dfrac{3,6}{40}=0,09(mol)\]
$c)$ Theo PTHH $(1);(2)→n_{HCl}=2.(n_{MgO}+n_{Mg})=2.(0,2+0,09)=0,58(mol)$
$\to V=\dfrac{0,58}{0,5}=0,0116(l)$
$d)$
Theo PTHH $(1);(2)→n_{MgCl_2}=n_{MgO}+n_{Mg}=0,2+0,09=0,29(mol)$
Theo PTHH $(3)→n_{Mg(OH)_2}=n_{MgCl_2}=0,29(mol)$
Theo PTHH $(4)→n_{MgO\;(4)}=n_{Mg(OH)_2}=0,29(mol)$
$\to a=0,29.40=11,6(g)$
