Giả sử mdd HCl = 100 (g) => nHCl = 0,8 (mol)
HCl + NaOH -> NaCl + H2O
0,8.......0,8...........0,8 (mol)
mdd NaOH = \(\frac{0,8.40}{40\%}=80\left(g\right)\)
mdd sau = 100+80=180(g)
\(C\%_{NaCl}=\frac{0,8.58,5}{180}.100\%=26\%\)
Gọi: x là khối lượng dd HCl
mHCl = 0.292x (g)
nHCl = x/125 mol
NaOH + HCl --> NaCl + H2O
x/125___x/125___x/125
mNaOH = 0.32x g
mdd NaOH = 0.8x g
mNaCl = 0.468x g
mdd sau phản ứng = x + 0.8x = 1.8x (g)
C%NaCl = 0.468x/1.8x *100% = 26%