a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)

c, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)

e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

Ta có: \(n_{MgO}=\dfrac{7}{40}=0,175\left(mol\right)\)

\(n_{H_2SO_4}=0,6.1=0,6\left(mol\right)\)

PT: \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

Xét tỉ lệ: \(\dfrac{0,175}{1}< \dfrac{0,6}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{MgSO_4}=n_{H_2SO_4\left(pư\right)}=n_{MgO}=0,175\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,6-0,175=0,425\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{MgSO_4}}=\dfrac{0,175}{0,6}=\dfrac{7}{24}\left(M\right)\\C_{M_{H_2SO_4}}=\dfrac{0,425}{0,6}=\dfrac{17}{24}\left(M\right)\end{matrix}\right.\)

\(n_{HCl}=1\cdot0,2=0,2\left(mol\right)\\ PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\\ a,n_{MgO}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\\ \Rightarrow m=m_{MgO}=0,1\cdot40=4\left(g\right)\\ b,n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\\ \Rightarrow m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\\ c,m_{CT_{HCl}}=0,2\cdot36,5=7,3\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{7,3}{250}\cdot100\%=2,92\%\)

\(n_{H_2O}=n_{MgO}=0,1\left(mol\right)\\ \Rightarrow m_{H_2O}=0,1\cdot18=1,8\left(g\right)\\ \Rightarrow m_{dd_{MgCl_2}}=4+250-1,8=252,2\left(g\right)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{9,5}{252,2}\cdot100\%\approx3,77\%\)

                                        Na₂O + H₂O → 2NaOH

                                           1            1            2

                                          0,1                       0,2

a).  n_Na2O=\(\dfrac{6,2}{62}\)= 0,1(mol)

       C_M=\(\dfrac{n}{V}\)=\(\dfrac{0,1}{4}\)= 0,025M

b).              Na₂O + H₂SO₄ → Na₂SO₄ + H₂O

                      1            1                1            1

                     0,1        0,1

m_H2SO4= n.M = 0,1 . 98 = 9,8g

⇒m_ddH2SO4= m_ct=\(\dfrac{mct.100\%}{C\%}\)= \(\dfrac{9,8.100}{20}\)= 49(g).

\(Mg + H_2SO_4 \to MgSO_4 + H_2\\ n_{MgSO_4}= n_{H_2} = n_{H_2SO_4} = n_{Mg} = \dfrac{4,8}{24} = 0,2(mol)\\ \Rightarrow m_{dd\ H_2SO_4} = \dfrac{0,2.98}{19,6\%} = 100(gam)\\ \Rightarrow m_{dd\ sau\ pư} = 4,8 + 100 - 0,2.2 = 104,4(gam)\\ \Rightarrow C\%_{MgSO_4} = \dfrac{0,2.120}{104,4}.100\% = 23\%\)

a, \(n_{Fe}=\frac{0.56}{56}=0.01\left(mol\right)\)

 \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

  0.01      0.01       0.01       0.01

\(V_{H_2}=0.01\times22.4=0.224\left(l\right)\)

b, \(m_{H_2SO_4}=0.01\times98=0.98\left(g\right)\)

   \(m_{ddH_2SO_4}=\frac{100\times0.98}{19.6}=5\left(g\right)\)

\(m_{FeSO_4}=0.01\times152=1.52\left(g\right)\)

\(C\%_{FeSO_4}=\frac{1.52\times100}{5}=30.4\%\)

Có: \(n_{NaOH}=0,2.1=0,2\left(mol\right)\)

PT: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

______0,2_____0,1_______0,1 (mol)

a, \(V_{ddH_2SO_4}=\dfrac{0,1}{2}=0,05\left(l\right)\)

b, \(C_{M_{Na_2SO_4}}=\dfrac{0,1}{0,2+0,05}=0,4M\)

Bạn tham khảo nhé!

Anh bổ sung câu c)

\(C_{MddNa_2SO_4}=\dfrac{0,25}{0,09879+0,5}=0,4175\left(M\right)\)