Ta có nH2 = \(\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Na + H2O \(\rightarrow\) NaOH + \(\dfrac{1}{2}\)H2
x........x............x.............x/2
K + H2O \(\rightarrow\) KOH + \(\dfrac{1}{2}\)H2
y.......y............y.............y/2
Bạn ơi trung hòa bằng dung dịch axít nào mới được chứ
Mình chọn HCl nha
NaOH + HCl \(\rightarrow\) NaCl + H2O
x..............x............x...........x
KOH + HCl \(\rightarrow\) KCl + H2O
y.............y.............y........y
=> \(\left\{{}\begin{matrix}\dfrac{x}{2}+\dfrac{y}{2}=0,1\\58,5x+74,5y=13,3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
=> mNa = 23 . 0,1 = 2,3 ( gam )
=> mK = 39 . 0,1 = 3,9 ( gam )
H2 + CuO \(\rightarrow\) Cu + H2O
0,1......0,1.......0,1........0,1
=> mCuO = 80 . 0,1 = 8 ( gam )
=> CM HCl = ( 0,1 + 0,1 ) : 0,4 = 0,5 M