2NaOH + 2HCl \(\rightarrow\)2NaCl + H2O (1)
2KOH + 2HCl \(\rightarrow\)2KCl + H2O (2)
Đặt nNaOH=a ; nKOH=b
Ta có:
\(\left\{{}\begin{matrix}40a+56b=3,04\\58,5a+74,5b=4,15\end{matrix}\right.\)
Giải hệ pt ta được:
a=0,02;b=0,04
mNaOH=0,02.40=0,8(g)
mKOH=56.0,04=2,24(g)
Gọi x;y là số mol NaOH;KOH
Theo gt:mhh=mNaOH+mKOH=40x+56y=3,04(1)
Ta có PTHH:
NaOH+HCl->NaCl+H2O(1)
x........................x.................(mol)
KOH+HCl->KCl+H2O(2)
y......................y.....................(mol)
Theo PTHH(1);(2):
mmuối=mNaCl+mKCl=58,5x+74,5y=4,15(2)
Giải phương trình(1);(2)=>\(\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,04\left(mol\right)\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}m_{NaOH}=40x=40.0,02=0,8\left(g\right)\\m_{KOH}=56y=56.0,04=2,24\left(g\right)\end{matrix}\right.\)
Vậy C%NaOH=\(\dfrac{0,8}{3,04}\).100%=26,3%
C%KOH=\(\dfrac{2,24}{3,04}\).100%=73,7%
Gọi x,y lần lượt là số mol của NaOH, KOH
\(NaOH+HCl\rightarrow NaCl+H_2O\) (1)
x ---------------\(\rightarrow\) x
\(KOH+HCl\rightarrow KCl+H_2O\) (2)
y---------------\(\rightarrow\) y
(1)(2)\(\Rightarrow\left\{{}\begin{matrix}40x+56y=3,04\\58,5x+74,5y=4,15\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,04\end{matrix}\right.\)
\(m_{NaOH}=0,02.40=0,8\left(g\right)\)
\(m_{KOH}=0,04.56=2,24\left(g\right)\)
