a, \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(n_{NaOH}=2n_{Na_2O}=0,2\left(mol\right)\Rightarrow C_{M_{NaOH}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)

\(a,n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,15    0,3           0,15      0,15

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

\(a,n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

PTHH :

\(Na_2O+H_2O\rightarrow2NaOH\)

0,1           0,1        0,2

\(C_{M\left(A\right)}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)

Đáp án A

\(m_{NaOH\left(bđ\right)}=\dfrac{90,7.8}{100}=7,256\left(g\right)\)

\(n_{Na_2O}=\dfrac{a}{62}\left(mol\right)\)

PTHH: Na₂O + H₂O --> 2NaOH

           \(\dfrac{a}{62}\)------------->\(\dfrac{a}{31}\)

=> \(m_{NaOH\left(sau.pư\right)}=\dfrac{a}{31}.40+7,256\left(g\right)\)

m_{dd sau pư} = a + 90,7 (g)

=> \(C\%_{dd.sau.pư}=\dfrac{\dfrac{40}{31}a+7,256}{a+90,7}.100\%=12\%\)

=> a = 3,1 (g)

Theo đề bài ta có :

nNa=\(\dfrac{a}{23}mol\)

Ta có PTHH : 1

\(2Na+2H2O\rightarrow2NaOH+H2\)
a/23.........a/23.........a/46

X=\(\dfrac{mct\left(mNaOH\right)}{m\text{dd}NaOH}.100\%=\dfrac{\left(\dfrac{a}{23}\right).40}{a+p-\left(\dfrac{a}{46}\right)}=\dfrac{40a}{23a+23p}\)(1)

Ta có : nNa2O=\(\dfrac{b}{62}mol\)

PTHH 2 :
\(Na2O+H2O\rightarrow2NaOH\: \)

b/62mol................2.(b/62)mol

=> X=\(\dfrac{mct}{m\text{dd}}.100\%=\dfrac{\left(2.\left(\dfrac{b}{62}\right)\right).40}{b+p}=\dfrac{40b}{124b+124p}\left(2\right)\)

Ta cho (1)=(2)
ta có biểu thức :
\(\dfrac{40a}{23a+23p}=\dfrac{40b}{124b+124p}\)

Bạn tự rút gọn biểu thức nhé!

CHO MÌNH CHỮA LẠI 1 CHÚT nãy ghi nhầm =.=

Theo đề bài ta có :

nNa=\(\dfrac{a}{23}mol\)

PTHH 1 :

\(2Na+2H2O\rightarrow2NaOH+H2\uparrow\)

a/23mol..................2(a/23)mol....1/2(a/23)mol

=> X=\(\dfrac{mct}{m\text{dd}}.100\%=\dfrac{40.2.\left(\dfrac{a}{23}\right)}{a+p-2.\dfrac{1}{2}\left(\dfrac{a}{23}\right)}\)=\(\dfrac{80a}{23a+23p}\)(1)

Theo đề bài ta có :

nNa2O=\(\dfrac{b}{62}mol\)

PTHH 2 :

\(Na2O+H2O\rightarrow2NaOH\)

b/62mol.................2(b/62)mol

=> X= \(\dfrac{mct}{m\text{đ}}.100\%=\dfrac{40.2\left(\dfrac{b}{62}\right)}{b+p}=\dfrac{80b}{62b+62p}\left(2\right)\)

Cho (1)=(2)

Ta có biểu thức : \(\dfrac{80a}{23a+23p}\)= \(\dfrac{80b}{62b+62p}\)

LÀM TẮT NHA

Dùng điện thoại khó ghi lắm .

Na +2H2O --> 2NaOH + H2 (1)

a/23 mol.................................a/23mol

Na2O + H2O --> 2NaOH(2)

b/62mol...............................b/32mol

mdd(1) = a+p-a/23.x%=40a/22a+23p

mdd(2) = b+ p.x%=40b /31b+31p

=>p=ab/b - 31p

Vậy....

đánh mất 30 phút v:

\(m_{NaOH\left(A\right)}=20.5\%=1\left(g\right)\)

Trong B:  

gọi x là khối lượng Na2O thêm vào , x>0 (g)

\(10\%=\dfrac{\dfrac{80}{62}x+1}{x+20}\)

\(\rightarrow x=0,84\left(g\right)\)

Vậy khối Na2O thêm vào dd A là 0,84 (g) 

b, \(m_{KOH\left(A\right)}=2\%.20=0,4\left(g\right)\)

\(C\%_{KOH\left(B\right)}=\dfrac{0,4}{20+0,84}.100\%=1,92\%\)

a, \(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{Na_2O}=\dfrac{3,1}{62}=0,05\left(mol\right)\)

Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,1\left(mol\right)\)

m _{dd sau pư} = 3,1 + 50 = 53,1 (g)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,1.40}{53,1}.100\%\approx7,53\%\)

b, \(2Na+2H_2O\rightarrow2NaOH+H_2\)

Ta có: \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{NaOH}=n_{Na}=0,2\left(mol\right)\\n_{H_2}=\dfrac{1}{2}n_{Na}=0,1\left(mol\right)\end{matrix}\right.\)

Ta có: m _{dd sau pư} = 4,6 + 95,6 - 0,1.2 = 100 (g)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,2.40}{100}.100\%=8\%\)

Gọi số mol NaOH là a (mol)

\(n_{Ba\left(OH\right)_2}=\dfrac{20,52}{171}=0,12\left(mol\right)\); \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Ta có sơ đồ:
\(21,9\left(g\right)X\left\{{}\begin{matrix}Na\\Ba\\Na_2O\\BaO\end{matrix}\right.+H_2O\rightarrow\left\{{}\begin{matrix}Ba\left(OH\right)_2:0,12\left(mol\right)\\NaOH:a\left(mol\right)\end{matrix}\right.+H_2:0,05\left(mol\right)\)

Bảo toàn H: \(n_{H_2O}=\dfrac{0,12.2+a+0,05.2}{2}=0,17+0,5a\left(mol\right)\)

Bảo toàn khối lượng:

\(m_X+m_{H_2O}=m_{Ba\left(OH\right)_2}+m_{NaOH}+m_{H_2O}\)

=> \(21,9+18\left(0,17+0,5a\right)=20,52+40a+0,05.2\)

=> a = 0,14 (mol)

=> m = 0,14.40 = 5,6 (g)

Na2O=0,5 mol

Na2O+H2O->2NaOH

0,5-----------------1 mol

ta có m NaOH=1.40+40=80g

=>C%=\(\dfrac{80}{431}100=18,561\%\)

Dd B chứa NaOH.

PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Ta có: \(n_{NaCl}=\dfrac{4,68}{58,5}=0,08\left(mol\right)\)

Theo PT: \(n_{NaOH\left(80\left(g\right)dd\right)}=n_{NaCl}=0,08\left(mol\right)\)

\(\Rightarrow n_{NaOH\left(200\left(g\right)dd\right)}=\dfrac{0,08.200}{80}=0,2\left(mol\right)\)

PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(\left\{{}\begin{matrix}23n_{Na}+62n_{Na_2O}=5,4\\n_{Na}+2n_{Na_2O}=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{Na}=0,1\left(mol\right)\\n_{Na_2O}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2}=\dfrac{1}{2}n_{Na}=0,05\left(mol\right)\)

Ta có: m _{dd B} = m_A + m_H2O - m_H2

⇒ 200 = 5,4 + m_H2O - 0,05.2

⇒ m_H2O = 194,7 (g)