n_H2 = 2.24/22.4 = 0.1 (mol) 

Na + H₂O => NaOH + 1/2 H₂

0.2....................0.2..........0.1

m_Na = 0.2 * 23 = 4.6 (g) 

m_Na2O = 17 - 4.6 = 12.4 (g) 

n_Na2O = 12.4/62 = 0.2 (mol) 

Na₂O + H₂O => 2NaOH 

0.2........................0.4

n_NaOH = 0.2 + 0.4 = 0.6 (mol) 

m_NaOH = 0.6 * 40 = 24 (g) 

n_CuO = 24/80 = 0.3 (mol) 

CuO + H₂ -t⁰-> Cu + H₂O

1...........1

0.3.........0.1

LTL : 0.3/1 > 0.1/1 

=> CuO dư 

n_Cu = n_H2 = 0.1 (mol) 

m_Cu = 0.1 * 64 = 6.4 (g) 

\(Đặt:n_{MnO_2}=a\left(mol\right),n_{KMnO_4}=b\left(mol\right)\)

\(m_{hh}=87a+158b=37.96\left(g\right)\left(1\right)\)

\(n_{Cl_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)

\(n_{Cl_2}=a+2.5b=0.45\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.4,b=0.02\)

\(\%MnO_2=\dfrac{0.4\cdot87}{37.96}\cdot100\%=91.68\%\\\%KMnO_4=100-91.68=8.32\% \)

\(m_M=m_{KCl}+m_{MnCl_2}=0.02\cdot74.5+\left(0.4+0.02\right)\cdot126=54.41g\)

a)

Ca + 2H₂O \(\rightarrow\) Ca(OH)₂ + H₂ (1)

Ca + H₂O \(\rightarrow\) Ca(OH)₂ (2)

T/d với nước dư => hỗn hợp hết

n_H2 = 3,36_/22,4 = 0,15(mol)

Theo PT(1) => n_H2 = n_Ca = 0,15(mol)

=> m_Ca = 0,15 . 40 = 6(g)

=> m_CaO = 17,2 - 6 =11,2(g)

b) n_CaO = 11,2_/56 = 0,2(mol)

Theo PT(1)(2) => tổng n_Ca(OH)2 = n_(Ca+CaO) = 0,15 + 0,2 = 0,35(mol)

=> m_Ca(OH)2 = 0, 35 . 74=25,9(g) = m_{ct trong dd sau pứ}

\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Na}=2.0,3=0,6\left(mol\right)\\ a,m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Na_2O}=26,2-13,8=12,4\left(g\right)\\b, n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ n_{NaOH\left(tổng\right)}=n_{Na}+2.n_{Na_2O}=0,6+\dfrac{12,4}{62}=0,8\left(mol\right)\\ m_{c.tan}=m_{NaOH}=0,8.40=32\left(g\right)\\ c,m_{ddNaOH}=m_{hh}+m_{H_2O}-m_{H_2}=26,2+200-0,3.2=225,6\left(g\right)\\ C\%_{ddNaOH}=\dfrac{32}{225,6}.100\approx14,185\%\)

Anh có làm rồi nha!

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ pthh:Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\) 
           0,3                                        0,3 
 \(m_{Ca}=0,3.40=12\left(g\right)\\ m_{CaO}=30-12=18\left(g\right)\)  
t cho Qùy tím vào dd 
Qùy tím hóa đỏ là axit 
Qùy tím hóa xanh là bazo 
 

\(a,n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: Ba + 2H₂O ---> Ba(OH)₂ + H₂

            0,3<-------------0,3<---------0,3

=> m_Ba = 0,3.137 = 41,1 (g)

=> m_K2O = 59,9 - 41,1 = 18,8 (g)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Ba}=\dfrac{41,1}{59,9}.100\%=68,61\%\\\%m_{K_2O}=100\%-68,61\%=31,39\%\end{matrix}\right.\)

\(b,n_{K_2O}=\dfrac{18,8}{94}=0,2\left(mol\right)\)

PTHH: K₂O + H₂O ---> 2KOH

          0,2----------------->0,4

Các chất tan trong dd sau phản ứng: KOH, Ba(OH)₂

\(\rightarrow\left\{{}\begin{matrix}m_{KOH}=0,4.56=22,4\left(g\right)\\m_{Ba\left(OH\right)_2}=0,3.171=51,3\left(g\right)\end{matrix}\right.\)

Ca     +    2H₂O   → Ca(OH)₂ + H₂↑     (1)

CaO  +     H₂O    → Ca(OH)₂                (2)

n_H2= 0,075 mol

Theo phương trình phản ứng (1), ta có: n_Ca=n_H2= 0,075 mol → m_Ca= 0,075 x 40=3 (g)

→m_CaO= m_{hỗn hợp} - m_CaO = 8,6 - 3 = 5,6 (g)

a, \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

m_{hh Zn và Fe} = 21,6-3 = 18,6 (g)

PTHH: Zn + H₂SO₄ → ZnSO₄ + H₂

Mol:      x                                           x

PTHH: Fe + H₂SO₄ → FeSO₄ + H₂

Mol:      y                                           y 

Ta có: \(\left\{{}\begin{matrix}65x+56y=18,6\\x+y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(\%m_{Zn}=\dfrac{0,2.65.100\%}{21,6}=60,19\%\)

\(\%m_{Fe}=\dfrac{0,1.56.100\%}{21,6}=25,93\%\)

\(\%m_{Cu}=100-60,19-25,93=13,88\%\)

b,

PTHH: Zn + H₂SO₄ → ZnSO₄ + H₂

Mol:     0,2     0,2                            0,2

PTHH: Fe + H₂SO₄ → FeSO₄ + H₂

Mol:     0,1      0,1                            0,1 

\(m_{H_2SO_4}=\left(0,1+0,2\right).98=29,4\left(g\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{29,4.100\%}{25\%}=117,6\left(g\right)\)

c,m_{dd sau pư} = 21,6+117,6- (0,1+0,2).2 = 138,6 (g)

\(C\%_{ddZnSO_4}=\dfrac{0,2.161.100\%}{138,6}=23,23\%\)

\(C\%_{ddFeSO_4}=\dfrac{0,1.152.100\%}{138,6}=10,97\%\)