Gọi \(\left\{{}\begin{matrix}n_{NaOH}=x\left(mol\right)\\n_{KOH}=y\left(mol\right)\end{matrix}\right.\)
PTHH: NaOH + HCl ---> NaCl + H2O
x------------------>x
KOH + HCl ---> KCl + H2O
y---------------->y
`=>` \(\left\{{}\begin{matrix}40x+56y=3,04\\58,5x+74,5x=4,15\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}x=0,02\\y=0,04\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}m_{NaOH}=0,02.40=0,8\left(g\right)\\m_{KOH}=0,04.74,5=2,98\left(g\right)\end{matrix}\right.\)
Gọi n NaOH là x
n KOH là y
NAOH + HCl---> NACl+ H2O(1)
x----------->x---------->x------>x(mol)
KOH+ HCl-------> KCl+ H2O(2)
y--------->y---------->y---------->y(mol)
Từ (1), (2) ta có hệ pt:
40x+ 56y=3,04
58,5x+ 74,5y= 4,15
Giải ra ta được:
x=0,02
y=0,04
=> nNaOH= 0,02(mol)
nKOH= 0,04(mol)
=> mNaOH= 0,02. 40=0,8(g)
mKOH=0,04. 56=2,24(g)
