\(m_{NaOH}=200.10\%=20\left(g\right)\Rightarrow n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
PTHH: NaOH + HCl ---> NaCl + H2O
0,5---->0,5-------->0,5
\(m_{HCl}=0,5.36,5=18,25\left(g\right)\Rightarrow m_{dd.HCl}=\dfrac{18,25}{3,65\%}=500\left(g\right)\)
mdd sau pứ = 500 + 200 = 700 (g)
=> \(C\%_{NaCl}=\dfrac{0,5.58,5}{700}.100\%=4,18\%\)
\(n_{NaOH}=\dfrac{10\%.200}{100\%.40}=0,5\left(mol\right)\)
Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)
0,5 0,5 0,5
\(n_{HCl}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)
\(m_{ddHCl}=\dfrac{18,25.100}{3,65}=500\left(g\right)\)
\(n_{NaCl}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{NaCl}=0,5.58,5=29,25\left(g\right)\)
\(m_{ddspu}=200+500=700\left(g\right)\)
\(C_{NaCl}=\dfrac{29,25.100}{700}=4,18\%\)
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