a) Gọi $n_{Fe} = a(mol) ; n_{MgO} = b(mol) \Rightarrow 56a + 40b = 23,3(1)$
$Fe + 2HCl \to FeCl_2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2O$

$\Rightarrow m_{muối} = 127a + 95b = 53,9(2)$

Từ (1)(2) suy ra a = \(\dfrac{23}{96}\) ; b = \(\dfrac{593}{2400}\)

\(\%m_{Fe}=\dfrac{\dfrac{23}{96}.56}{23,3}.100\%=57,58\%\)

$\%m_{MgO} = 100\% - 57,58\% = 42,42\%$

b) $n_{HCl} = 2a + 2b = \dfrac{73}{75}$

$m_{dd\ HCl} = \dfrac{ \dfrac{73}{75}.36,5}{7,3\%} = 486,67(gam)$

Hỗn hợp X gồm : $CO(a\ mol) ; CO_2(b\ mol)$

Ta có : \(M_X=\dfrac{28a+44b}{a+b}=20.2=40\)(g/mol)

mà : $n_{CO} = a + b = \dfrac{8,96}{22,4} = 0,4(mol)$
Suy ra :  a = 0,1; b = 0,3

$CO + O_{oxit} \to CO_2$
$n_{O\ trong\ oxit} = n_{CO_2} = 0,3(mol)$
$\Rightarrpw n_{Fe} =\dfrac{16 - 0,3.16}{56} = 0,2(mol)$

Ta có : 

$n_{Fe} : n_O = 0,2 : 0,3 = 2 : 3$. Do đó, oxit cần tìm là $Fe_2O_3$

vào gr đuy 🥺 

$n_{CuSO_4} = \dfrac{6,02.10^{23}}{6,02.10^{23}} = 1(mol)$
$m_{CuSO_4} = 1.160 = 160(gam)$

cảm ơn ạ

$n_{Mg} = \dfrac{4,8}{24} = 0,2(mol)$
$n_{HCl} = \dfrac{100.14,6\%}{36,5} = 0,4(mol)$

$Mg + 2HCl \to MgCl_2 + H_2$

Ta thấy : 

$n_{Mg} :1 = n_{HCl} :2$ nên phản ứng vừa đử.

$n_{H_2} = n_{Mg} = 0,2(mol)$
$\Rightarrow m_{dd\ sau\ pư} = 4,8 + 100 - 0,2.2 = 104,4(gam)$
$C\%_{MgCl_2} = \dfrac{0,2.95}{104,4}.100\% = 18,2\%$

a) $Zn + H_2SO_4 \to ZnSO_4 + H_2$
$n_{Zn} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$m_{Zn} = 0,2.65 = 13(gam)$
b) $n_{H_2SO_4} = n_{H_2} = 0,2(mol)$
$m_{H_2SO_4} = 0,2.98 = 19,6(gam)$
c) $n_{ZnSO_4} = n_{H_2} = 0,2(mol)$
$m_{ZnSO_4} = 0,2.161 = 32,2(gam)$

Em cảm ơn ạ 

Gọi $n_{KClO_3} = a(mol) ; n_{KMnO_4} = b(mol) \Rightarrow 122,5a + 158b = 87,7(1)$

TH1 : 

$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
$2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$
$m_{O_2} = 87,7 - 71,7 = 16(gam)$
$\Rightarrow n_{O_2} = 1,5a + 0,5b = \dfrac{16}{32} = 0,5(2)$

Từ (1)(2) suy ra a = 0,2 ; b = 0,4

$\%m_{KClO_3} = \dfrac{0,2.122,5}{87,7}.100\% = 27,93\%$

$\%m_{KMnO_4} = 100\% -27,93\% = 72,07\%$

b) $m_{O_2} =  87,7 - 81,3 = 6,4(gam) \Rightarrow n_{O_2} = 0,2(mol)$
$n_{Cl_2} = 1,15(mol)$
Bảo toàn electron : 

$6n_{KClO_3} + 5n_{KMnO_4} = 4n_{O_2} + 2n_{Cl_2}$
$\Rightarrow 6a + 5b = 0,2.4 + 1,15.2(3)$

Từ (1)(3) suy ra a = \(\dfrac{513}{3355}\) ; b = \(\dfrac{2929}{6710}\)

\(\%m_{KClO_3}=\dfrac{\dfrac{513}{3355}.122,5}{87,7}.100\%=21,36\%\)

$\%m_{KMnO_4} = 100\% - 21,36\% = 78,64\%$

$n_{O_2} =\dfrac{67,2}{22,4}.\dfrac{1}{5} = 0,6(mol)$
$n_C = \dfrac{6}{12} = 0,5(mol)$

$C + O_2 \xrightarrow{t^o} CO_2$
Ta thấy : 

$n_C : 1 < n_{O_2} :1$ nên $O_2$ dư.

Do đó, Cacbon có cháy hết.

$1) 2Al(NO_3)_3 \xrightarrow{t^o} Al_2O_3 + 6NO_2 + \dfrac{3}{2}O_2\\ 4CO + Fe_3O_4 \xrightarrow{t^o} 3Fe + 4CO_2\\ 3) Al_2O_3 + 3K_2SO_4 \to Al_2(SO_4)_3 + 3K_2SO_4 + 3H_2O$

1) \(4Al\left(NO_3\right)_3\xrightarrow[CaO]{t^o}2Al_2O_3+12NO_2+3O_2\)

2) \(4CO+Fe_3O_4\xrightarrow[]{t^o}3Fe+4CO_2\)

3) \(Al_2O_3+6KHSO_4\rightarrow3K_2SO_4+Al_2\left(SO_4\right)_3+3H_2O\)