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) 

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

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

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

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ pthh:Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\) 
           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 
 

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

Anh có làm rồi nha!

Mg+2HCl->MgCl₂+H₂

x------------------------x

Zn+2HCl->ZnCl₂+H₂

y-------------------------y

Ta có :

\(\left\{{}\begin{matrix}24x+65y=17,8\\x+y=\dfrac{8,96}{22,4}\end{matrix}\right.\)

=>x=0,2 mol, y=0,2 mol

=>m_Mg=0,2.24=4,8g

=>m _Zn=0,2.65=13g

=>m _HCl=0,8.36,5=29,2g

Ba + 2H2O --> Ba(OH)2 + H2

BaO + H2O --> Ba(OH)2

nH2= 0.56/22.4=0.025 (mol)

=> nBa= 0.025 (mol)

mBa= 0.025*137=3.425g

mBaO= 6.485-3.425=3.06g

nBaO= 0.02 (mol)

%Ba= 3.425/6.485*100%= 52.81%

%BaO= 100 - 52.81= 47.19%

b) nBa(OH)2= 0.025+0.002= 0.045 (mol)

mBa(OH)2 = 0.045*171=7.695g