Na2O+H2O->2NaOH

0,129----------------0,258

n Na2O=0,129 mol

=>n NaOH=0,258+0,5=0,758mol

=>CM =\(\dfrac{0,758}{0,5}\)=1,516M

Na2O+H2O->2NaOH

0,129----------------0,258

n Na2O=0,129 mol

=>n NaOH=0,258+0,5=0,758mol

=>CM =\(\dfrac{0,758}{0,5}\)=1,516M

4P+5O2-to>2P2O5

0,4---0,5-----0,2

n P=0,4 mol

=>VO2=0,5.22,4=11,2l

=>Vkk=1,12.5=5,6l

a) \(m_{CuSO_4}=\dfrac{5.200}{100}=10\left(g\right)\)

b) Gọi khối lượng CuSO₄ là a (g)

Có: \(C\%=\dfrac{a}{a+190}.100\%=5\%\Rightarrow a=10\left(g\right)\)

c) 

\(n_{Na_2SO_3}=0,2.1=0,2\left(mol\right)\Rightarrow m_{Na_2SO_3}=0,2.126=25,2\left(g\right)\)

a) 4P + 5O₂ --t^o--> 2P₂O₅

b) \(n_P=\dfrac{15,5}{31}=0,5\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

           0,5--------------->0,25

=> m_P2O5 = 0,25.142 = 35,5 (g)

c) 

PTHH: P₂O₅ + 3H₂O --> 2H₃PO₄

              0,25------------>0,5

=> \(C\%=\dfrac{0,5.98}{100}.100\%=49\%\)

4P+5O2-to>2P2O5

0,5-------------0,25

n P=0,5 mol

=>m P2O5=0,25.142=35,5g

c)P2O5+3H2O->2H3PO4

    0,25---------------------0,5

=>C%=\(\dfrac{0,5.98}{100}100=49g\)

1) S + O₂ --t^o--> SO₂

2) 2Al + 6HCl --> 2AlCl₃ + 3H₂

3) 2Na + 2H₂O --> 2NaOH + H₂

4) 2KNO₃ --t^o--> 2KNO₂ + O₂

5) 2KClO₃ --t^o--> 2KCl + 3O₂

6) BaO + H₂O --> Ba(OH)₂
7) Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O

8) Fe_xO_y + yCO --t^o--> xFe + yCO₂

9) Ba + 2H₂O --> Ba(OH)₂ + H₂

10) 2SO₂ + O₂ --t^o,V₂O₅--> 2SO₃

\(S+O_2\underrightarrow{t^o}SO_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\\ 2KNO_3\underrightarrow{t^o}2KNO_2+O_2\\ 2KClO_3\xrightarrow[xtMnO_2]{t^o}2KCl+3O_2\\ BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ Fe_xO_y+yCO\underrightarrow{t^o}xFe+yCO_2\\ H_2O+Ba\rightarrow Ba\left(OH\right)_2\\ 2SO_2+O_2\xrightarrow[xtV_2O_5]{t^O}2SO_3\)

\(n_{CaO}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PTHH: CaCO₃ --t^o--> CaO + CO₂

           0,15<-----------0,15

=> m = 0,15.100 = 15 (g)

Anh kudo giải hóa hoài z a

\(n_{FeS}=\dfrac{22}{88}=0,25\left(mol\right)\)

PTHH: Fe + S --t^o--> FeS

        0,25<------------0,25

=> m = 0,25.56 = 14 (g)

`a)PTHH:`

`Fe + S` $\xrightarrow[]{t^o}$ `FeS`

`0,25`            `0,25`                         `(mol)`

`n_[FeS] = 22 / 88 = 0,25 (mol)`

`b) m_[Fe] = 0,25 . 56 = 14 (g)`

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

          0,1---->0,1----------------->0,1

=> \(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\end{matrix}\right.\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O
LTL: 0,25 > 0,1 => CuO dư

Theo pthh: n_Cu = n_H2 = 0,1 (mol)

=> m_Cu = 0,1.64 = 6,4 (g)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
         0,1    0,2                           0,2 
\(V_{H_2}=0,2.22,4=4,48l\\ C_M=\dfrac{0,2}{0,2}=1M\\ n_{CuO}=\dfrac{20}{80}=0,25\left(G\right)\\ pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ LTL:0,25>0,1\) 
=>CuO dư 
\(n_{Cu}=n_{H_2}=0,1\left(mol\right)\\ m_{Cu}=0,1.64=6,4g\)

2Na+Br2-to,xt>2NaBr

thank pác 

(1) 2Fe + 3Cl₂ --t^o--> 2FeCl₃

(2) FeCl₃ + 3NaOH --> Fe(OH)₃ + 3NaCl

(3) 2Fe(OH)₃ --t^o--> Fe₂O₃ + 3H₂O

(4) Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O