\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\\ Đặt:n_{Na}=a\left(mol\right);n_{Ba}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}23a+137b=18,3\\0,5.22,4a+22,4b=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ a,\%m_{Na}=\dfrac{0,2.23}{18,3}.100\approx25,137\%\\ \Rightarrow\%m_{Ba}\approx74,863\%\\ m_{ddB}=m_{hhA}+m_{H_2O}-m_{H_2}=18,3+102,1-0,2.2=120\left(g\right)\\ n_{NaOH}=n_{Na}=0,2\left(mol\right);n_{Ba\left(OH\right)_2}=n_{Ba}=0,1\left(mol\right)\\ b,C\%_{ddNaOH}=\dfrac{0,2.40}{120}.100\approx6,667\%\\ C\%_{ddBa\left(OH\right)_2}=\dfrac{0,1.171}{120}.100=14,25\%\)
đặt nNa=x , nBa=y
mX=23x+137y=18,3g (1)
nH2=0,5a+b=\(\dfrac{4,48}{22,4}=0,2mol\) (2)
Từ 1 và 2 ta có : x=0,2 , y=0,1 mol
=>mNa=0,2.23=4,6g
=>mBa=0,1.127=13,7g
Ta có pt:
2Na+2H2O->2NaOH+H2
0,2---------------0,2
2Ba+2H2O->Ba(OH)2+H2
0,1-------------0,05
ta có :
C%NaOH=\(\dfrac{0,2.40}{18,3+102,1}.100=6,64\%\)
C% Ba(OH)2=\(\dfrac{0,05.171}{18,3+102,1}.100=7,1\%\)
