a, \(n_{CO_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

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

c, \(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)

Ta có: \(n_{CO_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

PT: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

Theo PT: \(n_{NaOH}=2n_{CO_2}=0,4\left(mol\right)\)

\(\Rightarrow a=C\%_{NaOH}=\dfrac{0,4.40}{200}.100\%=8\%\)

Ta có: \(n_{Fe_2O_3}=\dfrac{1,6}{160}=0,01\left(mol\right)\)

PT: \(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)

Theo PT: \(n_{Fe}=2n_{Fe_2O_3}=0,02\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,02.56=1,12\left(g\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

\(\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)

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

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}\left(M\right)\)

Ta có: m_BaCl2 = 0,2.208 = 41,6 (g)

\(\Rightarrow C\%_{BaCl_2}=\dfrac{41,6}{41,6+150}.100\%\approx21,7\%\)

a, \(2Mg+O_2\underrightarrow{t^o}2MgO\)

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

⇒ m_MgO = 0,4.40 = 16 (g)

c, \(n_{MgO}=\dfrac{20}{40}=0,5\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{MgO}=0,25\left(mol\right)\)

⇒ V_O2 = 0,25.22,4 = 5,6 (l)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

Theo PT: \(n_{Al_2O_3\left(LT\right)}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)

Mà: H% = 60%

\(\Rightarrow n_{Al_2O_3\left(TT\right)}=0,1.60\%=0,06\left(mol\right)\)

⇒ m_{Al2O3 (TT)} = 0,06.102 = 6,12 (g)

Câu 1:

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

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: n_Fe = n_H2 = 0,1 (mol)

⇒ m_Fe = 0,1.56 = 5,6 (g)

⇒ m_Cu = 12 - 5,6 = 6,4 (g)

Câu 2:

Ta có: \(n_{H_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

Theo PT: \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2.24}{15,6}.100\%\approx30,8\%\\\%m_{Ag}\approx69,2\%\end{matrix}\right.\)

Câu 3:

\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

Theo PT: \(\left\{{}\begin{matrix}n_{Al\left(LT\right)}=2n_{Al_2O_3}=0,2\left(mol\right)\\n_{O_2\left(LT\right)}=\dfrac{3}{2}n_{Al_2O_3}=0,15\left(mol\right)\end{matrix}\right.\)

Mà: H% = 90% \(\Rightarrow\left\{{}\begin{matrix}n_{Al\left(TT\right)}=0,2.90\%=0,18\left(mol\right)\\n_{O_2\left(TT\right)}=0,15.90\%=0,135\left(mol\right)\end{matrix}\right.\)

⇒ m_Al (TT) = 0,18.27 = 4,86 (g)

V_{O2 (TT)} = 0,135.24,79 = 3,34665 (l)

Có lẽ đề cho khí H₂ ở đktc chứ không phải đkc bạn nhỉ?

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

Theo PT: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ m _{dd sau pư} = 0,2.27 + 300 - 0,3.2 = 304,8 (g)

\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342}{304,8}.100\%\approx11,2\%\)

\(n_{Na_2CO_3}=\dfrac{12,72}{106}=0,12\left(mol\right)\)

PT: \(Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl\)

Theo PT: n_CaCO3 = n_Na2CO3 = 0,12 (mol)

⇒ m = m_CaCO3 = 0,12.100 = 12 (g)