đề j kì v bn
NaOH hay NaCl
làm theo NaOH
ta có: mNaOH= 200. 6%= 12( g)
\(\Rightarrow\) nNaOH= \(\dfrac{12}{40}\)= 0,3( mol)
PTPU
HCl+ NaOH\(\rightarrow\) NaCl+ H2O
0,3......0,3..........0,3............... mol
\(\Rightarrow\) C%HCl= \(\dfrac{0,3.36,5}{200}\). 100%= 5,475%
ta có: mdd sau pư= mdd NaOH+ mdd HCl
= 200+ 200= 400( g)
\(\Rightarrow\) C%NaCl= \(\dfrac{0,3.58,5}{400}\). 100%= 4,3875%
đính chính NaCl = NaOH
mNaOH = \(\dfrac{6.200}{100}\)= 12 g => nNaOH = \(\dfrac{12}{40}\)= 0,3 mol
HCl + NaOH -> NaCl + H2O
0,3<--0,3----->0,3 mol
=> C%HCl = \(\dfrac{0,3.36,5}{200}\).100% = 5,475 %
=>C%NaCl = \(\dfrac{0,3.58,5}{12+200}.100\%\) = 8,278%
