Cu không pư với dd H₂SO₄ loãng.

Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)

\(\Rightarrow m_{Cu}=10-6,5=3,5\left(g\right)\)

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

PT: \(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

Theo PT: \(n_{Cl_2}=\dfrac{3}{2}n_{Fe}=0,15\left(mol\right)\)

\(\Rightarrow V_{Cl_2}=0,15.24,79=3,7185\left(l\right)\)

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

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

               0,1       ->                       0,1 (mol)

\(\rightarrow V_{H_{2\left(đktc\right)}}=0,1.22,4=2,24\left(l\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)

\(Fe\left(OH\right)_2\underrightarrow{t^o}FeO+H_2O\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

\(FeSO_4+2NaOH\rightarrow Na_2SO_4+Fe\left(OH\right)_2\)

\(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

\(FeCl_3+3NaCl\rightarrow3NaCl+Fe\left(OH\right)_2\)

\(2Fe\left(OH\right)_2\underrightarrow{t^o}Fe_2O_3+3H_2O\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

\(Fe_2\left(SO_4\right)_3+6NaOH\rightarrow3Na_2SO_4+2Fe\left(OH\right)_3\)

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1mol\\ CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\\ n_{BaCO_3}=n_{CO_2}=0,1mol\\ Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\\ n_{H_2SO_4}=n_{Ba\left(OH\right)_2}=0,1mol\\ m_{ddH_2SO_4}=\dfrac{0,1.98}{20\%}\cdot100\%=49g\\ V_{ddH_2SO_4}=\dfrac{49}{1,14}=42,98ml\)

\(BaCl_2+Na_2SO_4\rightarrow BaSO_4+2NaCl\)

\(n_{H_2SO_4}=0,1.0,75=0,075mol\\ H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)

0,075           0,15               0,075          0,15

\(a)m_{K_2SO_4}=0,075.175=13,05mol\)

\(b)H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ n_{NaOH}=0,075.2=0,15mol\\ m_{ddNaOH}=\dfrac{0,15.40}{15\%}\cdot100\%=40g\\ V_{ddNaOH}=\dfrac{40}{1,05}=38,1ml\)      

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

a, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)

c, PT: \(HCl+KOH\rightarrow KCl+H_2O\)

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

\(\Rightarrow m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400\left(g\right)\)

\(\Rightarrow V_{ddKOH}=\dfrac{400}{1,045}\approx382,78\left(ml\right)\)

\(H_2SO_4+BaCl_2\xrightarrow[]{}BaSO_4+2HCl\)