a,\(m_{HCl}=120.36,5\%=43,8\left(g\right)\Rightarrow n_{HCl}=\dfrac{43,8}{36,5}=1,2\left(mol\right)\)

PTHH: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

Mol:        x           6x          

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

Mol:         y           6y          

Ta có:\(\left\{{}\begin{matrix}102x+160y=26,2\\6x+6y=1,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

\(\%m_{Al_2O_3}=\dfrac{0,1.102.100\%}{26,2}=38,93\%;\%m_{Fe_2O_3}=100\%-38,93\%=61,07\%\)

câu c tương tự câu a

\(n_{Zn}=\dfrac{2,6}{65}=0,04\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:    0,04   0,08                  0,04

\(\Rightarrow C_{M_{ddHCl}}=\dfrac{0,08}{0,2}=0,4M\)

  \(V_{H_2}=0,04.22,4=0,896\left(l\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

\(\Rightarrow n_{HCl}=2n_{Zn}=2.\dfrac{2,6}{65}=0,08\left(mol\right)\)

\(\Rightarrow C_{M\left(HCl\right)}=\dfrac{0,08}{0,2}=0,4M\)

\(n_{H_2}=n_{Zn}=0,08\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,08.22,4=1,792\left(l\right)\)

a,\(n_{SO_3}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)

PTHH: 2NaOH + SO₃ → Na₂SO₄ + H₂O

Mol:       0,25       0,125     0,125

\(\Rightarrow m_{Na_2SO_4}=0,125.142=17,75\left(g\right)\)

b,m_NaOH = 0,25.40 = 10 (g)

 \(C\%_{ddNaOH}=\dfrac{10.100\%}{480}=2,08\%\)

c,m_{dd sau pứ} = 480 + 0,125.80 = 490 (g)

\(C\%_{ddNa_2SO_4}=\dfrac{17,75.100\%}{490}=3,62\%\)

Ta có: \(n_{CO_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)

Bảo toàn Cacbon: \(n_{CaCO_3}=n_{CO_2}=0,02\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CaCO_3}=\dfrac{0,02\cdot100}{5}\cdot100\%=40\%\\\%m_{CaSO_4}=60\%\end{matrix}\right.\)

Ta có: \(n_{NaOH}=0,3\cdot2=0,6\left(mol\right)\)

Bảo toàn Clo và Natri: \(n_{NaOH}=n_{NaCl}=n_{HCl}=0,6\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\left(M\right)\\m_{NaCl}=0,6\cdot58,5=35,1\left(g\right)\end{matrix}\right.\)

NaOH + HCl --> NaCl+H2O (1)

2NaOH +H2SO4 --> Na2SO4 +2H2O (2)

Đặt n_{NaOH (1)}= a(mol)

n_NaOH(2) = b (mol)

=>a + b = 0,3.2 =0,6( *)

Theo PT (1) : n_NaCl = n_NaOH(1) = a(mol)

Theo PT (2) : n_Na2SO4=12

 n_NaOH(2) = 0,5b(mol)

=>58,5a + 71b =40,1(**)

Từ (*), (**) => a= 0,2 ; b = 0,4 

 n_HCl = n_{NaOH (1)}=0,2 mol 

=> x=CMHCl=0,20,2=1M

n_H2SO4 = n_{NaOH (2)}=0,4 mol 

y=CMH2SO4=0,40,2=2M

a,\(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

PTHH: Na₂O + H₂O → 2NaOH

Mol:      0,1                      0,2

\(\Rightarrow C_{M_{ddA}}=\dfrac{0,2}{0,1}=2M\)

b,m_ddKOH = 6,2+100.1=106,2 (g) 

 \(\Rightarrow D_{ddKOH}=\dfrac{106,2}{100}=1,062\left(g/cm^3\right)\)

c,m_KOH = 0,2.56 = 11,2 (g)

\(C\%_{ddKOH}=\dfrac{11,2.100\%}{106,2}=10,55\%\)

??? Sao lại KOH vậy 

c_m=\(\dfrac{n}{V}\)

>_<

Đổi: \(500ml=0,5l\)

\(C_{M_{HCl}}=\dfrac{n_{HCl}}{V_{HCl}}\Rightarrow n_{HCl}=C_{M_{HCl}}.V_{HCl}=2.0,5=1\left(mol\right)\)

\(n_{HCl}=0,5\cdot2=1\left(mol\right)\)

a,\(n_{HCl}=0,2.1=0,2\left(mol\right)\)

PTHH: HCl + KOH → KCl + H₂O

Mol:      0,2     0,2         0,2

b,\(V_{ddKOH}=\dfrac{0,2}{2}=0,1\left(l\right)=100\left(ml\right)\)

c,V_{dd sau pứ} = 200 + 100 = 300(ml) 

d,m_KCl = 0,2.74,5 = 14,9 (g)

Tên muối là kali clorua