a, \(Na_2O+H_2O\rightarrow2NaOH\)

\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

\(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)

b, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

\(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)

a, \(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)

\(\Rightarrow CM_{NaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\)

b, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)

a) 

\(n_{Na_2O}=\dfrac{31}{62}=0,5\left(mol\right)\)

PTHH: Na₂O + H₂O --> 2NaOH

______0,5--------------->1

=> \(C_{M\left(NaOH\right)}=\dfrac{1}{0,5}=2M\)

b) 

PTHH: H₂SO₄ + 2NaOH --> Na₂SO₄ + 2H₂O

______0,5<---------1

=> m_H2SO4 = 0,5.98 = 49(g)

=> \(m_{dd\left(H_2SO_4\right)}=\dfrac{49.100}{20}=245\left(g\right)\)

=> \(V_{dd\left(H_2SO_4\right)}=\dfrac{245}{1,14}=214,912\left(ml\right)\)

\(n_{Na_2O}=\dfrac{31}{62}=0,5(mol)\\ a,Na_2O+H_2O\to 2NaOH\\ \Rightarrow n_{NaOH}=1(mol)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{1}{0,5}=2M\\ b,2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ \Rightarrow n_{H_2SO_4}=0,5(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,5.98}{20\%}=245(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{245}{1,14}=214,91(ml)\)

nK₂O = \(\dfrac{23,5}{94}=0,25\left(mol\right)\)

Pt: K₂O + H₂O --> 2KOH

.....0,25 mol-------> 0,5 mol

C_{M KOH} = \(\dfrac{0,5}{0,5}=1\left(M\right)\)

b) Pt: 2KOH + H₂SO₄ --> K₂SO₄ + 2H₂O

..........0,5 mol->0,25 mol

mdd H2SO4 cần = \(\dfrac{0,25.98}{20}.100=122,5\left(g\right)\)

c) nCuCl2 = 1 . 0,1 = 0,1 mol

Pt: CuCl2 + 2NaOH --> Cu(OH)2 + 2NaCl

.....0,1 mol---------------> 0,1 mol

Xét tỉ lệ mol giữa CuCl2 và NaOH:

\(\dfrac{0,1}{1}< \dfrac{0,5}{2}\)

Vậy NaOH dư

mCu(OH)2 thu được = 0,1 . 98 = 9,8 (g)

a)

$Na_2O + H_2O \to 2NaOH$

n Na2O = 15,5/62 = 0,25(mol)

n NaOH = 2n Na2O = 0,5(mol)

=> CM NaOH = 0,5/0,5 = 1M

b) $2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$

n H2SO4 = 1/2 n NaOH = 0,25(mol)

=> m dd H2SO4 = 0,25.98/20% = 122,5(gam)

=> V dd H2SO4 = m / D = 122,5/1,14 =107,46(ml)

c) n Na2SO4 = n H2SO4 = 0,25(mol)

CM Na2SO4 = 0,25/0,10746 = 2,33M

a)

`\(Na_2O++H_{2_{ }}O->2NaOH\)

\(n_{Na_2O}=\dfrac{15,5}{62}=0,25mol\)

\(n_{Na_2O}=2n_{Na_2O}=0,5mol\)

\(C_{M_{NaOH}}=\dfrac{0,5}{0,5}\)=1M

BaO+H2O -> Ba(OH)2 
0,02             0,02  
a) CM = n/V = 0,02/0,02 = 1M
b) Ba(OH)2 + H2SO4 -> BaSO4 +2H2O
      0,02          0,02
=> m = 0,392 g 
D = m/V = 1,14
=> 0,392/V = 1,14 => V = 0,34l

cảm ơn bạn nhá

n_baco3=\(\dfrac{m}{M}=\dfrac{10}{100}=0,1\left(mol\right)\)

gọi x, y lần lượt là số mol của CuO và PbO

pthh: CuO + CO \(\rightarrow\) Cu + CO2

x . ..... ...... . ........ ............x

PbO + CO \(\rightarrow\) Pb + CO2

y ...... ...... ....... ........y

CO2 + Ca(OH)2 \(\rightarrow\) CaCO3 + H2O

Theo pthh: n_CO2=n_CaCO3=0,1(mol)

Theo đề bài: m_{2 oxit}=4(g)

\(\Leftrightarrow80x+223y=4\)

và n_CO2=0,1(mol)

\(\Leftrightarrow\) x + y = 0,1

Ta có hệ pt:

\(\Rightarrow\left\{{}\begin{matrix}80x+223y=4\\x+y=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,12797\left(mol\right)\\y=-0,02797\left(mol\right)\end{matrix}\right.\)

Sai đề rồi

Bài 1 :

a, Ta có n_Na2O = \(\dfrac{15,5}{62}\) = 0,25 ( mol )

Na₂O + H₂O \(\rightarrow\) 2NaOH

0,25.......0,25.........0,5

=> C_{M NaOH} = n : V = 0,5 : 0,5 = 1M