\(a) CuO + 2HCl \to CuCl_2 + H_2O\\ b) n_{CuO} = \dfrac{4,8}{80} = 0,06(mol) ; n_{HCl} = \dfrac{100.3,65\%}{36,5} = 0,1(mol)\\ \dfrac{n_{CuO}}{1}= 0,06 > \dfrac{n_{HCl}}{2} = 0,05 \to CuO\ dư\\ n_{CuCl_2} = n_{CuO\ pư} = \dfrac{1}{2}n_{HCl} = 0,05(mol)\\ m_{dd\ sau\ pư} = 0,05.80 + 100 = 104(gam)\\ C\%_{CuCl_2} = \dfrac{0,05.135}{104}.100\% = 6,49\%\)

Bài 2L

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, Ta có: 24n_Mg + 27n_Al = 5,1 (g) (1)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(m_{HCl}=100.21,9\%=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

Có: n_{HCl (pư)} = 2n_H2 = 0,5 (mol) < 0,6 → HCl dư.

⇒ n_{HCl (dư)} = 0,6 - 0,5 = 0,1 (mol)

Theo PT: \(n_{H_2}=n_{Mg}+\dfrac{3}{2}n_{Al}=0,25\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,1\left(mol\right)\\n_{Al}=0,1\left(mol\right)\end{matrix}\right.\)

Ta có: m _{dd sau pư} = 5,1 + 100 - 0,25.2 = 104,6 (g)

Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{104,6}.100\%\approx3,49\%\\C\%_{MgCl_2}=\dfrac{0,1.95}{104,6}.100\%\approx9,08\%\\C\%_{AlCl_3}=\dfrac{0,1.133,5}{104,6}.100\%\approx12,76\%\end{matrix}\right.\)

Bài 3:

Gọi: n_H2 = a (mol)

BTNT H, có: n_HCl = 2n_H2 = 2a (mol)

Theo ĐLBT KL, có: m_KL + m_HCl = m_A + m_H2

⇒ 5 + 2a.36,5 = 5,71 + 2a ⇒ a = 0,01 (mol)

⇒ V_H2 = 0,01.22,4 = 0,224 (l)

Bài 1:

Ta có: \(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)

PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

Theo PT: \(n_{CuCl_2}=n_{Cu}=0,02\left(mol\right)\Rightarrow m_{CuCl_2}=0,02.135=2,7\left(g\right)\)

a. Ta có: \(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)

Ta lại có: \(C_{\%_{HCl}}=\dfrac{m_{ct_{HCl}}}{100}.100\%=7,3\%\)

=> m_HCl = 7,3(g)

=> \(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

PTHH:

Fe₃O₄ + 8HCl ---> FeCl₂ + 2FeCl₃ + 4H₂O

    1   --->   8

    0,1 --->  0,2

=> \(\dfrac{0,1}{1}>\dfrac{0,2}{8}\)

Vậy Fe₃O₄ dư

=> m_dư = 23,2 - 7,3 = 15,9 (g)

b. Theo PT: \(n_{FeCl_2}=\dfrac{1}{8}.n_{HCl}=\dfrac{1}{8}.0,2=0,025\left(mol\right)\)

=> \(m_{FeCl_2}=0,025.127=3,175\left(g\right)\)

Theo PT: \(n_{FeCl_3}=\dfrac{1}{4}.n_{HCl}=\dfrac{1}{4}.0,2=0,05\left(mol\right)\)

=> \(m_{FeCl_3}=0,05.162,5=8,125\left(g\right)\)

=> \(m_{muối}=8,125+3,175=11,3\left(g\right)\)

c. Ta có: m_{dung dịch sau PỨ} = \(23,2+100=123,2\left(g\right)\)

Theo PT: \(n_{H_2O}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)

=> \(m_{H_2O}=0,1.18=1,8\left(g\right)\)

m_{các chất sau PỨ} = 1,8 + 11,3 = 13,1(g)

=> \(C_{\%_{sauPỨ}}=\dfrac{13,1}{123,2}.100\%=10,63\%\)

tùy theo nồng độ các chất có trong dung dịch nhé

- ko có nồng độ phần trăm sao tính đc

a, Ta có: \(n_{CaCO_3}=\dfrac{300}{100}=3\left(mol\right)\)

\(m_{HCl}=400.7,3\%=29,2\left(g\right)\Rightarrow n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\)

PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

Xét tỉ lệ: \(\dfrac{3}{1}>\dfrac{0,8}{2}\), ta được CaCO₃ dư.

Theo PT: \(n_{CO_2}=\dfrac{1}{2}n_{HCl}=0,4\left(mol\right)\Rightarrow V_{H_2}=0,4.22,4=8,96\left(l\right)\)

b, Theo PT: \(n_{CaCO_3\left(pư\right)}=n_{CaCl_2}=\dfrac{1}{2}n_{HCl}=0,4\left(mol\right)\)

Ta có: m _{dd sau pư} = 0,4.100 + 400 - 0,4.44 = 422,4 (g)

\(\Rightarrow C\%_{CaCl_2}=\dfrac{0,4.111}{422,4}.100\%\approx10,51\%\)

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

Bài 7:

Ta có: \(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)

\(m_{H_2SO_4}=100.40\%=40\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{40}{98}=\dfrac{20}{49}\left(mol\right)\)

PT: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{\dfrac{20}{49}}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{ZnSO_4}=n_{H_2SO_4\left(pư\right)}=n_{ZnO}=0,2\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=\dfrac{20}{49}-0,2=\dfrac{51}{245}\left(mol\right)\)

Ta có: m _{dd sau pư} = 16,2 + 100 = 116,2 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnSO_4}=\dfrac{0,2.161}{116,2}.100\%\approx27,71\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{\dfrac{51}{245}.98}{116,2}.100\%\approx17,56\%\end{matrix}\right.\)

Bài 8:

Gọi oxit cần tìm là AO.

Ta có: \(m_{HCl}=10.21,9\%=2,19\left(g\right)\Rightarrow n_{HCl}=\dfrac{2,19}{36,5}=0,06\left(mol\right)\)

PT: \(AO+2HCl\rightarrow ACl_2+H_2O\)

Theo PT: \(n_{AO}=\dfrac{1}{2}n_{HCl}=0,03\left(mol\right)\)

\(\Rightarrow M_{AO}=\dfrac{2,4}{0,03}=80\left(g/mol\right)\)

\(\Rightarrow M_A+16=80\Rightarrow M_A=64\left(g/mol\right)\)

→ A là Cu.

Vậy: Đó là oxit của đồng.

Ta có: \(m_{Ba\left(OH\right)_2}=100.17,1\%=17,1\left(g\right)\Rightarrow n_{Ba\left(OH\right)_2}=\dfrac{17,1}{171}=0,1\left(mol\right)\)

\(m_{H_2SO_4}=150.9,8\%=14,7\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)

PT: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_{4\downarrow}+2H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{BaSO_4}=n_{Ba\left(OH\right)_2}=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,15-0,1=0,05\left(mol\right)\)

Ta có: m _{dd sau pư} = 100 + 150 - 0,1.233 = 226,7 (g)

\(\Rightarrow C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,05.98}{226,7}.100\%\approx2,16\%\)

\(n_{HCl}=\dfrac{100.14,6\%}{36,5}=0,4\left(mol\right)\\ n_{MgCO_3}=\dfrac{50}{84}=\dfrac{25}{42}\left(mol\right)\\ PTHH:MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\\ Vì:0,4:2< \dfrac{25}{42}:1\\ \Rightarrow MgCO_3dư\\ \Rightarrow ddsau:MgCl_2\\n_{MgCO_3\left(p.ứ\right)}=n_{CO_2}= n_{MgCl_2}=\dfrac{n_{HCl}}{2}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ m_{ddsau}=m_{MgCO_3\left(p.ứ\right)}+m_{ddHCl}-m_{CO_2}=0,2.84+100-0,2.44=108\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,2.95}{108}.100\approx17,593\%\%\)