\(n_{HCl}=\dfrac{150.7,3\%}{36,5}=0,3\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\\ a,n_{Al}=n_{AlCl_3}=\dfrac{2}{6}.0,3=0,1\left(mol\right)\\ \Rightarrow m_{Al}=0,1.27=2,7\left(g\right)\\ b,m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\ c,n_{H_2}=\dfrac{3}{6}.0,3=0,15\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)

nHCl=150.7,3%36,5=0,3(mol)2Al+6HCl→2AlCl3+3H2↑a,nAl=nAlCl3=26.0,3=0,1(mol)⇒mAl=0,1.27=2,7(g)b,mAlCl3=0,1.133,5=13,35(g)c,nH2=

Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)

a. PTHH: Mg + H₂SO₄ ---> MgSO₄ + H₂↑

Theo PT: \(n_{H_2}=n_{Mg}=0,4\left(mol\right)\)

=> \(V_{H_2}=0,4.22,4=8,96\left(lít\right)\)

b. Theo PT: \(n_{H_2SO_4}=n_{Mg}=0,4\left(mol\right)\)

=> \(m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{39,2}{m_{dd_{H_2SO_4}}}.100\%=10\%\)

=> \(m_{dd_{H_2SO_4}}=392\left(g\right)\)

c. Ta có: \(m_{H_2}=0,4.2=0,8\left(g\right)\)

=> \(m_{dd_{MgSO_4}}=9,6+392-0,8=400,8\left(g\right)\)

Theo PT: \(n_{MgSO_4}=n_{Mg}=0,4\left(mol\right)\)

=> \(m_{MgSO_4}=0,4.120=48\left(g\right)\)

=> \(C_{\%_{MgSO_4}}=\dfrac{48}{400,8}.100\%=11,98\%\)

\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{Mg}=\dfrac{4,8}{24}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

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

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

0,2        0,4           0,2            0,2

\(V_{H_2}=\left(0,1+0,2\right).22,4=6,72l\\ b)V_{ddHCl}=\dfrac{0,2+0,4}{2}=0,3l\\ c)m_{muối}=0,1.127+95.0,2=31,7g\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\\C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\end{matrix}\right.\)

Bài 1 :

PTHH : Zn + H₂SO₄ ------> ZnSO₄ + H₂

\(n_{Zn}=\frac{m}{M}=\frac{26}{65}=0,4\left(mol\right)\)

Theo PTHH : n_H2 = n_Zn = 0,4 mol

=> Khối lượng H₂ được tạo ra ở đktc là :

\(V=n\times22,4\)

\(\Rightarrow V_{H_2}=0,4\times22,4\)

\(\Rightarrow V_{H_2}=8,96\left(l\right)\)

Theo PTHH : n_ZnSO4 = n_Zn = 0,4 mol

=> Khối lượng muối được tạo thành là :

\(m=n\times M\)

\(\Rightarrow m_{ZnSO_4}=0,4\times161\)

\(\Rightarrow m_{ZnSO_4}=64,4\left(g\right)\)

b) Theo PT : \(n_{H_2SO_4}=n_{Zn}=0,4\left(mol\right)\)

=> Khối lượng \(H_2SO_4\)cần dùng cho phản ứng là :

\(m=n\times M\)

\(\Rightarrow m_{H_2SO_4}=0,4\times98\)

\(\Rightarrow m_{H_2SO_4}=39,2\left(g\right)\)

c) Nồng độ phần trăm thu được sau phản ứng là :

\(C\%=\frac{m_{ct}}{m_{dd}}\times100\%\)

\(\Rightarrow C\%=\frac{39,2}{64,4}\times100\%\approx60,9\%\)

Vậy :.........................

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a.Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{ddHCl}=\dfrac{0,4.36,5.100}{7,3}=200\left(g\right)\\ c.m_{ddsau}=4,8+200-0,2.2=204,4\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,2.95}{204,4}.100\approx9,295\%\\ d.V_{ddHCl}=\dfrac{200}{1,05}=\dfrac{4000}{21}\left(ml\right)=\dfrac{4}{21}\left(l\right)\\ C_{MddHCl}=\dfrac{0,4}{\dfrac{4}{21}}=2,1\left(M\right)\)

a) \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: CaCO₃ + 2HCl ---> CaCl₂ + CO₂ + H₂O

             0,3<-----0,6<------------------0,3

=> m = 0,3.100 = 30 (g)

b) \(C\%_{HCl}=\dfrac{0,6.36,5}{150}.100\%=14,6\%\)

a) K₂CO₃ + 2HCl --> 2KCl + CO₂ + H₂O

b) \(n_{K_2CO_3}=\dfrac{13,8}{138}=0,1\left(mol\right)\)

PTHH: K₂CO₃ + 2HCl --> 2KCl + CO₂ + H₂O

______0,1----->0,2------>0,2--->0,1

=> V_CO2 = 0,1.22,4 = 2,24 (l)

c) m_HCl = 0,2.36,5 = 7,3 (g)

\(m_{ddHCl}=\dfrac{7,3.100}{7,3}=100\left(g\right)\)

d) m_KCl = 0,2.74,5 = 14,9 (g)

m_{dd sau pư} = 13,8 + 100 - 0,1.44 = 109,4 (g)

=> \(C\%\left(KCl\right)=\dfrac{14,9}{109,4}.100\%=13,62\%\)