`2K+2H_2O->2KOH+H_2` (sp: Kali hidroxit, Hidro)

0,2---------------------------0,1 mol

`K_2O+H_2O->2KOH`

`n_(H_2)=((2,24)/(22,4))=0,1 mol`

`->m_(K)=0,2.39=7,8g`

`->m_(K_2O)=17,2-7,8=9,4g`

\(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PTHH:

2K + 2H₂O ---> 2KOH + H₂

0,2<---------------0,2<----0,1

=> \(\left\{{}\begin{matrix}m_K=0,2.39=7,8\left(g\right)\\m_{K_2O}=12,5-7,8=4,7\left(g\right)\end{matrix}\right.\)

\(n_{K_2O}=\dfrac{4,7}{94}=0,05\left(mol\right)\)

PTHH: K₂O + H₂O ---> 2KOH

            0,05-------------->0,1

=> m_KOH = (0,2 + 0,1).56 = 16,8 (g)

2K+2H2O->2KOH+H2

0,2--------------0,2-------0,1mol

K2O+H2O->2KOH

0,05--------------0,1

n H2=0,1 mol

=>m K=0,2.39=7,8g

=>m K2O=4,7g=>n K2O=0,05 mol

=>m bazo=0,3.56=16,8g

a. PTHH 1 : Ba + H₂O -> Ba(OH)₂ + H₂

                   0,025                             0,025

PTHH 2 : BaO + H₂O -> Ba(OH)₂ 

\(n_{H_2}=\dfrac{0.56}{22,4}=0,025\left(mol\right)\)

\(m_{Ba}=0,025.137=3,425\left(g\right)\)

\(m_{BaO}=6,485-3,425=3,06\left(g\right)\)

\(n_{BaO}=\dfrac{3.06}{153}=0,02\left(mol\right)\)

b. \(\%m_{BaO}=\dfrac{3,06}{6,485}.100=47,2\%\)

\(\%m_{Ba}=100\%-47,2\%=52,8\%\)

giúp minh với các bạn 

a, Cu không tác dụng với dd HCl.

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)

\(\Rightarrow m_{Cu}=19,4-13=6,4\left(g\right)\)

c, Ta có: \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{19,4}.100\%\approx67,01\%\\\%m_{Cu}\approx32,99\%\end{matrix}\right.\)

Bạn tham khảo nhé!

ta có Cu không tác dụng với HCl nên lượng khí thoát ra là của Mg

pthh  Mg  +  2HCl  => MgCl2  + H2

nH2  = 4.48 / 22,4 =0,2 mol mà nMg = nH2 =0,2 

=> mMg= 0,2 * 24 =4,8g

mCu = 10 - 4,8 =5,2 g

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

Theo PTHH: \(n_{H_2}=n_{Mg}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,2\cdot24=4,8\left(g\right)\)

\(\Rightarrow m_{Cu}=10-4,8=5,2\left(g\right)\)

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

Anh có làm rồi nha!

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

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