Sửa đề: 9,2 gam Na

\(a,n_{Na_2O}=\dfrac{9,2}{23}=0,4\left(mol\right)\)

PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

            0,4------------------>0,8

\(\rightarrow C_{M\left(NaOH\right)}=\dfrac{0,8}{0,5}=1,6M\)

\(b,n_{K_2O}=\dfrac{37,6}{94}=0,4\left(mol\right)\)

PTHH: \(K_2O+H_2O\rightarrow2KOH\)

            0,4----------------->0,8

\(\rightarrow C\%_{KOH}=\dfrac{0,8.56}{362,4+37,6}.100\%=11,2\%\)

\(1.Na_2O+H_2O\rightarrow2NaOH\\ 2.m_{rắn}=m_{CuO}=6,9g\\ m_{Na_2O}=10-6,9=3,1g\\ n_{Na_2O}=\dfrac{3,1}{62}=0,05mol\\ n_{NaOH}=0,05.2=0,1mol\\ 200ml=0,2l\\ C_{M_X}=C_{M_{NaOH}}=\dfrac{0,1}{0,2}=0,5M\)

$n_{Na_2O} = \dfrac{18,6}{62} = 0,3(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,6(mol)$
$C\%_{NaOH} = \dfrac{0,6.40}{200}.100\% = 12\%$

PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{18,6}{62}=0,6\left(mol\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,6\cdot40}{18,2+200}\cdot100\%\approx11\%\)

Đáp án : C.

Xét Na : n H 2 = 1 2 n N a = a 46 m o l

Bảo toàn khối lượng :

m_{dd sau} = m_Na + m H 2 O – m H 2  = a + p – 2.  a 46 (g)

=> C%_NaOH = x %

Xét Na₂O

=> m_{dd sau} = m N a 2 O + m H 2 O = b + p

nNa = 6.9 : 23 = 0.3 mol

           4Na + O2 ->2 Na2O

mol :  0.3 ->           0.15

          Na2O + H2O -> 2NaOH

mol : 0.15 ->                0.3

mdd = 0.15 x 62 + 140.7 = 150g

C% NaOH = 0.3x40: 150 x 100% = 8%

\(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=0,2.2=0,4\left(mol\right)\\ C_{MddNaOH}=\dfrac{0,4}{0,4}=1\left(M\right)\)

\(n_{Na_2O}=\dfrac{12,4}{62}=0,2mol\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=0,2.2=0,4mol\\ C_{M_{NaOH}}=\dfrac{0,4}{0,4}=1M\)

\(n_{Na_2O}=\dfrac{12,4}{62}=0,2mol\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,2  \(\rightarrow\)    0,2   \(\rightarrow\)   0,4

\(C_{M_{NaOH}}=\dfrac{0,4}{\dfrac{500}{1000}}=0,8M\)