\(m_{NaOH}=\dfrac{200\cdot10\%}{100\%}=20g\) \(\Rightarrow n_{NaOH}=0,5mol\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,25   \(\leftarrow\)  0,25  \(\leftarrow\)  0,5

\(m_{Na_2O}=0,25\cdot62=15,5g\)

\(m_{H_2O}=0,25\cdot18=4,5g\)

\(Na_2O+H_2O\rightarrow2NaOH\)

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

BTKL : 

\(m_{H_2O}=m_{NaOH}-m_{Na_2O}=8-6.2=1.8\left(g\right)\)

$n_{Ba} = n_{Ba(OH)_2} = 0,12(mol)$

$n_{H_2} = \dfrac{1,12}{22,4} = 0,05(mol)$
Gọi $n_{Na} = a ; n_O = b$

Ta có : 

$23a + 16b + 0,12.137 = 21,1$

Bảo toàn electron : $a + 0,12.2 = 2b + 0,05.2$

Suy ra $a = \dfrac{177}{1550} ; b = \dfrac{197}{1550}$

Suy ra $m_{NaOH} = \dfrac{177}{1550}.40 = 4,57(gam)$

\(n_{Na_2O}=\dfrac{6.2}{62}=0.1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(0.1.........................0.2\)

\(C_{M_{NaOH}}=\dfrac{0.2}{0.5}=0.4\left(M\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

\(0.2.............0.2\)

\(m_{HCl}=0.2\cdot36.5=7.3\left(g\right)\)

Chúc em học tốt !!!

Chọn C

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

PTHH: Na₂O + H₂O --> 2NaOH

             0,1-------------->0,2

=> m_NaOH = 0,2.40 = 8 (g)

m_dd = 6,2 + 93,8 = 100 (g)

=> \(C\%=\dfrac{8}{100}.100\%=8\%\)

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.

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

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

Ta có pthh

Na2O + H2O2NaOH

Áp dụng định luật bảo toàn khối lượng ta có

mNa2O+mH2O=mNaOH

mNaOH=6,2+93,8=100 g

Ta có

nH2O=6,262=0,1mol6,262=0,1���

Theo pthh

nNa2O=

Vậy nồng độ của dd tạo thành là 8%

\(a,n_{Na_2O}=\dfrac{6,2}{62}=0,1mol\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=0,1.2=0,2mol\\ C_{\%A}=C_{\%NaOH}=\dfrac{0,2.40}{6,2+93,8}\cdot100\%=4\%\\ b.n_{CuSO_4}=\dfrac{200.16\%}{100\%.160}=0,2mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ \Rightarrow\dfrac{0,2}{1}>\dfrac{0,2}{2}\Rightarrow CuSO_4.dư\\ n_{Na_2SO_4}=n_{Cu\left(OH\right)_2}=n_{CuSO_4,pư}=0,2:2=0,1mol\\ m_{ddA}=6,2+93,8+200-0,1.98=290,2g\\ C_{\%Na_2SO_4}=\dfrac{0,1.142}{290,2}\cdot100\%\approx4,98\%\\ C_{\%CuSO_4,dư}=\dfrac{\left(0,2-0,1\right).160}{290,2}\cdot100\%\approx5,51\%\)