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

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

Chọn A

n N a 2 O = m N a 2 O M N a 2 O = 6,2 62 = 0,1   m o l

nNa2O=0.1(mol)

pthh:Na2O+H2O->2NaOH

theo pthh:nH2O=nNa2O->nH2O=0.1(mol)->mH2O=0.1*18=1.8(g)

b)Theo pthh:nNaOH=2 nNa2O

->nNaOH=0.1*2=0.2(mol)

CM=0.2:0.2=1(M)

PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\)

            \(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=n_{Na}+2n_{Na_2O}=\dfrac{4,6}{23}+2\cdot\dfrac{6,2}{62}=0,3\left(mol\right)\\n_{H_2}=\dfrac{1}{2}n_{Na}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,3\cdot40=12\left(g\right)\\m_{H_2}=0,05\cdot2=0,1\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{Na}+m_{Na_2O}+m_{H_2O}-m_{H_2}=110,7\left(g\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{12}{110,7}\cdot100\%\approx10,84\%\)

Theo gt ta có: $n_{Na}=0,2(mol);n_{Na_2O}=0,1(mol)$

$2Na+2H_2O\rightarrow 2NaOH+H_2$

$Na_2O+H_2O\rightarrow 2NaOH$

Ta có: $n_{NaOH}=0,4(mol);n_{H_2}=0,1(mol)$

Bảo toàn khối lượng ta có: $m_{dd}=110,6(g)$

$\Rightarrow \%C_{NaOH}=14,46\%$

\(n_{Na}=\dfrac{m}{M}=0,2mol\)

\(n_{Na2O}=\dfrac{m}{M}=0,1\left(mol\right)\)

\(BTNT\left(Na\right):n_{NaOH}=0,2+2.0,1=0,4\left(mol\right)\)

\(\Rightarrow m_{NaOH}=n.M=16g\)

\(PTHH:Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

...............0,2...................................0,1.

\(\Rightarrow m_{dd}=m_{KL}+m_{H2O}-m_{H2}=110,6g\)

\(\Rightarrow C\%_{NaOH}=\dfrac{m}{m_{dd}}.100\%=14,46\%\)

\(n_{Na_2O}=\dfrac{6,2}{62}=0,1mol\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=0,1.2=0,2mol\\ C_{M_X}=C_{M_{NaOH}}=\dfrac{0,2}{2}=0,1M\)

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

PT: \(Na_2O+H_2O\rightarrow2NaOH\)

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

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,2}{2}=0,1\left(m\right)\)

0,1M

0, 1M

\(a.Na_2O+H_2O\rightarrow2NaOH\\ b.n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\\ n_{NaOH}=2n_{Na_2O}=0,2\left(mol\right)\\ \Rightarrow CM_{NaOH}=\dfrac{0,2}{0,5}=0,4M\\ c.H_2SO_4+2NaOH\rightarrow Na_2SO_4+H_2O\\ n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,1\left(mol\right)\\ m_{ddH_2SO_4}=\dfrac{0,1.98}{9,8\%}=100\left(g\right)\)

1)
Na2O + H2O --> 2NaOH
0.1 0.2 (mol)
C% =(0,2. 40)/(100 + 6,2) = 7,53%