a) PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{6,2}{62}=0,2\left(mol\right)\) \(\Rightarrow C_{M_{NaOH}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)

b) PTHH: \(2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{CuSO_4}=0,2\cdot2,5=0,5\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,5}{1}\) \(\Rightarrow\) CuSO₄ còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{Cu\left(OH\right)_2}=0,1\left(mol\right)=n_{Na_2SO_4}\\n_{CuSO_4\left(dư\right)}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Cu\left(OH\right)_2}=0,1\cdot98=9,8\left(g\right)\\C_{M_{Na_2SO_4}}=\dfrac{0,1}{0,4+0,2}\approx0,17\left(M\right)\\C_{M_{CuSO_4\left(dư\right)}}=\dfrac{0,4}{0,6}\approx0,67\left(M\right)\end{matrix}\right.\)

Ta có : n_Na \(=\dfrac{4,6}{23}=0,2\left(mol\right)\)

PTHH : \(2Na+2H_2O\rightarrow2NaOH+H_2\uparrow\)

a) Theo pt :\(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b) Theo pt : \(n_{NaOH}=n_{Na}=0,2\left(mol\right)\)

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

\(m_{ddspu}=4,6+100-0,1.2=104,4\left(g\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{8}{104,4}.100\%=7,66\%\)

Mg+2HCl->MgCl2+H2

0,1---0,2-----0,1-----0,1

n Mg=0,1 mol

=>VH2=0,1.22,4=2,24l

C% HCl dư=\(\dfrac{0,2.36,5}{100}100\)=7,3%

=>C%MgCl2=\(\dfrac{0,1.95}{2,4+100-0,1.2}100=9,29\%\)

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

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

0,1  <    0,4                               ( mol )

0,1         0,2           0,1          0,1      ( mol )

\(V_{H_2}=0,1.22,4=2,24l\)

\(C\%_{HCl}=\dfrac{0,2.36,5}{100}.100\%=7,3\%\)

\(C\%_{MgCl_2}=\dfrac{0,1.95}{2,4+100-0,1.2}.100\%=9,29\%\)

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
           0,1        0,2          0,1        0,1 
\(V_{H_2\left(\text{đ}ktc\right)}=0,1.22,4=2,24\left(L\right)\\ V_{\text{đ}kt}=0,1.24=2,4l\\ V_{H_2\left(\text{đ}kc\right)}=0,1.24,79=2,479l\)
\(m_{\text{dd}}=2,4+100-0,2=102,2\left(g\right)\\ m_{MgCl_2}=95.0,1=9,5g\\ C\%=\dfrac{9,5}{102,2}.100\%=9,3\%\)
\(C\%_{HCl\left(d\right)}=\dfrac{0,2.36,5}{100}.100=7,3\%\)

\(n_{NaOH}=\dfrac{100.14,4\%}{100\%.40}=0,36mol\\ T=\dfrac{0,36}{0,28}\\ \Rightarrow1< T< 2\)
Phản ứng tạo 2 muối
\(n_{Na_2CO_3}=a;n_{NaHCO_3}=b\\ 2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\\ NaOH+CO_2\rightarrow NaHCO_3\\ \Rightarrow\left\{{}\begin{matrix}2a+b=0,36\\a+b=0,28\end{matrix}\right.\\ \Rightarrow a=0,08;b=0,2\\ C_{\%Na_2CO_3}=\dfrac{0,08.106}{0,28.44+100}\cdot100\%=7,55\%\\ C_{\%NaHCO_3}=\dfrac{0,2.84}{0,28.44+100}\cdot100\%=14,96\%\%\)

Cứu

\(C1:\\ n_{NaOH}=1.0,1=0,1mol\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaOH}=0,1mol\\ C_{M\left(NaOH\right)}=\dfrac{0,1}{0,1}=1M\\ C2:\\ n_{Na_2O}=\dfrac{6,2}{62}=0,1mol\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=2n_{Na_2O}=0,2mol\\ C_{\%NaOH}=\dfrac{0,2.40}{100+6,2}\cdot100=7,53\%\)

\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\\ PTHH:Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=2.0,25=0,5\left(mol\right)\\ a,C_{MddNaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\\ b,2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4}=n_{Na_2SO_4}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{H_2SO_4}=0,25.98=24,5\left(g\right)\\ m_{ddH_2SO_4}=\dfrac{24,5.100}{20}=122,5\left(g\right)\\ V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,456\left(ml\right)\\ c,V_{ddsau}=V_{ddNaOH}+V_{ddH_2SO_4}\approx0,5+0,107456=0,607456\left(l\right)\\C_{MddNa_2SO_4}\approx\dfrac{ 0,25}{0,607456}\approx0,411552\left(M\right)\)

a, \(Na_2O+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

b, Số mol \(H_2SO_4\) là: \(n_1=V.C_M=0,5.0,5=0,25\) (mol)

Số mol \(Na_2SO_4\) là \(n_2=\dfrac{28,4}{142}=0,2\) (mol)

Do \(n_2< n_1\) nên \(H_2SO_4\) còn dư

Suy ra số mol \(Na_2O\) tham gia phản ứng là: \(n=n_2=0,2\) (mol)

Khối lượng là: \(m_{Na_2O}=0,2.62=12,4g\)