\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

         1         2              1           1

       0,1        0,2

a) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(C_{M_{ddHCl}}=\dfrac{0,2}{1,5}=0,13\left(M\right)\)

b) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)

               1            1              1           1

              0,2          0,2

\(n_{NaOH}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(m_{NaOH}=0,2.40=8\left(g\right)\)

\(m_{ddNaOH}=\dfrac{8.100}{5}=160\left(g\right)\)

 Chúc bạn học tốt

\(a,n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{Mg}=0,2\left(mol\right)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,2}{1,5}=\dfrac{2}{15}M\\ b,n_{HCl}=\dfrac{2}{15}\cdot0,75=0,1\left(mol\right)\\ PTHH:HCl+NaOH\rightarrow NaCl+H_2O\\ \Rightarrow n_{NaOH}=n_{HCl}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{NaOH}}=0,1\cdot40=4\left(g\right)\\ \Rightarrow m_{dd_{NaOH}}=\dfrac{4\cdot100\%}{5\%}=80\left(g\right)\)

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{HCl}=2.n_{Mg}=2.0,1=0,2\left(mol\right)\\ a,C_{M\text{dd}HCl}=\dfrac{0,2}{1,5}=\dfrac{2}{15}\left(M\right)\\ b,n_{HCl}=\dfrac{0,75}{1,5}.0,2=0,1\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{HCl}=0,1\left(mol\right)\\ m_{\text{dd}NaOH}=\dfrac{0,1.40.100}{5}=80\left(g\right)\)

a) \(n_{Na}=\dfrac{11,5}{23}=0,5\left(mol\right)\)

\(n_{NaOH}=\dfrac{8\%.500}{40}=1\left(mol\right)\)

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

0,5---------------->0,5------->0,25

\(\Sigma n_{NaOH}=0,5+1=1,5\left(mol\right)\)

\(m_{ddsaupu}=11,5+500-0,25.2=511\left(g\right)\)

=> \(C\%_{NaOH}=\dfrac{1,5.40}{511}.100=11,74\%\)

b) Gọi thể tích dung dịch X cần tìm là V

 \(n_{H^+}=V.1+V.0,5.1=2V\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

Ta có : \(n_{H^+}=n_{OH^-}=1,5\left(mol\right)\)

=> 2V=1,5

=> V=0,75(lít)

C% sau PƯ=\(\dfrac{150.10\%+250.25\%}{150+250}.100\%=19,375\%\)

\(n_{Mg}=\dfrac{7.2}{24}=0.3\left(mol\right)\)

\(n_{HCl}=\dfrac{200\cdot14.6\%}{36.5}=0.8\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0.3.........0.6.........0.3..........0.3\)

\(n_{HCl\left(dư\right)}=0.8-0.6=0.2\left(mol\right)\)

\(KOH+HCl\rightarrow KCl+H_2O\)

\(0.2........0.2\)

\(V_{dd_{KOH}}=\dfrac{0.2}{2}=0.1\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng}}=7.2+200-0.3\cdot2=206.6\left(g\right)\)

\(m_{MgCl_2}=0.3\cdot95=28.5\left(g\right)\)

\(C\%MgCl_2=\dfrac{28.5}{206.6}\cdot100\%=13.8\%\)

\(C\%HCl\left(dư\right)=\dfrac{0.2\cdot36.5}{206.6}\cdot100\%=3.53\%\)

\(n_{HCl}=0,08.2=0,16\left(mol\right)\)

PTHH:

\(NaOH+HCl\rightarrow NaCl+H_2O\)

0,16          0,16       0,16

\(C_{M\left(NaOH\right)}=\dfrac{0,16}{0,025}=6,4\left(M\right)\)