Bài 3 :
a, \(NaOH+HCl\rightarrow NaCl+H_2O\)
b, \(m_{NaOH}=\frac{40.20}{100}=8\left(g\right)\)
\(\rightarrow n_{NaOH}=0,2\left(mol\right)\)
Theo pt: nHCl= nNaOH= 0,2 mol
\(\rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(\rightarrow m_{dd}=29,2\left(g\right)\)
c, \(m_{NaCl}=0,2.\left(23+35,5\right)=11,7\left(g\right)\)
\(\rightarrow m_{dd}=29,2+40=69,2\left(g\right)\)
\(\rightarrow\%_{NaCl}=\frac{11,7}{69,2}=16,9\%\)