\(PTHH:2Na+2H_2O\rightarrow2NaOH+H_2\)

\(n_{Na}=\dfrac{46}{23}=2\left(mol\right)\)

Theo pthh:

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

\(\Rightarrow\left\{{}\begin{matrix}m_{H_2}=1.2=2\left(g\right)\\m_{NaOH}=2.40=80\left(g\right)\end{matrix}\right.\)

Theo đlbtkl, ta có:

\(m_{ddNaOH}=m_{Na}+m_{H_2O}-m_{H_2}\)

\(=46+180-2=224\left(g\right)\)

\(C\%ddNaOH=\dfrac{80}{224}.100\approx35,7\%\)

2Na + 2H₂O → 2NaOH + H₂

\(m_{Na}=2\times23=46\left(g\right)\)

\(m_{H_2O}=200\times1=200\left(g\right)\)

Theo pT: \(n_{H_2}=\frac{1}{2}n_{Na}=\frac{1}{2}\times2=1\left(mol\right)\)

\(\Rightarrow m_{H_2}=1\times2=2\left(g\right)\)

Ta có: \(m_{ddA}=46+200-2=244\left(g\right)\)

Theo pT: \(n_{NaOH}=n_{Na}=2\left(mol\right)\)

\(\Rightarrow m_{NaOH}=2\times40=80\left(g\right)\)

\(\Rightarrow C\%_{NaOH}=\frac{80}{244}\times100\%=32,79\%\)

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

Theo PTHH (1) : \(n_{Na}=2mol;n_{NaOH}=2mol;n_{H_2}=1mol\)

Có : \(m_{H_2O}=V_{H_2O}.D_{H_2O}=200.1=200g\)

Vì \(n_{Na}=2< n_{H_2O}=\frac{200}{18}\) : Sau p/ứng (1) thì H₂O dư

Khối lượng dung dịch thu được :

\(m_{dd}=m_{Na}+m_{H_2O}-m_{H_2}=2.23+200-1.2=244g\)

Từ (1) => \(n_{NaOH}=2mol\Rightarrow m_{NaOH}=2.40=80g\)

Vậy \(C\%_{NaOH}=\frac{80}{244}.100=32,79\%\)

\(n_{K_2O}=\dfrac{21,15}{94}=0,225\left(mol\right)\\ pthh:K_2O+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\) 
         0,225                   0,225        0,1125 
\(m_{\text{dd}}=21,15+178,85-\left(0,1125.2\right)=199,775g\\ C\%=\dfrac{0,225.56}{199,775}.100\%=6,3\%\) 
c)
\(m_{\text{dd}}=21,15+\left(50+178,85\right)-\left(0,1125.2\right)=249,775g\\ C\%=\dfrac{0,225.56}{249,775}.100\%=5\%\)