Câu 1:

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

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

\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)

c, \(n_{HCl}=2n_{Fe}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{1}=0,4\left(l\right)=400\left(ml\right)\)

Câu 2:

a, \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

b, Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

Theo PT: \(n_{Al_2\left(SO_4\right)_3\left(LT\right)}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3\left(LT\right)}=0,2.342=68,4\left(g\right)\)

Mà: m_{Al2(SO4)3 (TT)} = 34,2 (g)

\(\Rightarrow H\%=\dfrac{34,2}{68,4}.100\%=50\%\)

Ta có: \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

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

Theo PT: n_Mg = n_H2 = 0,3 (mol)

⇒ m_Mg = 0,3.24 = 7,2 (g)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{7,2}{15}.100\%=48\%\\\%m_{Cu}=100-48=52\%\end{matrix}\right.\)

Ta có: \(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)

PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

_____0,4_______0,4______0,4 (mol)

a, m_H2SO4 = 0,4.98 = 39,2 (g)

b, m_CuSO4 = 0,4.160 = 64 (g)

Ta có: \(n_{O_2}=\dfrac{0,4958}{24,79}=0,02\left(mol\right)\)

PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Theo PT: \(n_{KMnO_4\left(LT\right)}=2n_{O_2}=0,04\left(mol\right)\)

Mà: H% = 80% ⇒ n_{KMnO4 (TT)} = 0,04:80% = 0,05 (mol)

⇒ m_{KMnO4 (TT)} = 0,05.158 = 7,9 (g)

a, Ta có: d_X/H2 = 75 ⇒ M_X = 75.2 = 150 (g/mol)

b, Gọi: CTPT của X là C_xH_yO_z

\(\Rightarrow x:y:z=\dfrac{54}{12}:\dfrac{5}{1}:\dfrac{16}{16}=9:10:2\)

→ X có CTPT dạng (C₉H₁₀O₂)_n

\(\Rightarrow n=\dfrac{150}{12.9+1.10+2.16}=1\)

Vậy: CTPT của X là C₉H₁₀O₂.

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Ta có: \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,3\left(mol\right)\)

\(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

⇒ m_Zn = 0,3.65 = 19,5 (g)

m_HCl = 0,6.36,5 = 21,9 (g)

c, m_ZnCl2 = 0,3.136 = 40,8 (g)

1 tấn = 1000 kg

a, Ta có: \(n_{CaCO_3}=\dfrac{1000}{100}=10\left(kmol\right)\)

PT: \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

______10______10_____10 (kmol)

⇒ m_CaO = 10.56 = 560 (kg)

m_CO2 = 10.44 = 440 (kg)

b, 504 (g) = 0,504 (kg)

\(\Rightarrow H\%=\dfrac{0,504}{560}.100\%=0,09\%\)

Ta có: \(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)

PT: \(S+O_2\underrightarrow{t^o}SO_2\)

Theo PT: \(n_{SO_2\left(LT\right)}=n_S=0,1\left(mol\right)\)

Mà: H% = 80%

⇒ n_{SO2 (TT)} = 0,1.80% = 0,08 (mol)

⇒ m_SO2 (TT) = 0,08.64 = 5,12 (g)

a, \(2Cu+O_2\underrightarrow{t^o}2CuO\)

b, \(n_{Cu}=\dfrac{3,2}{64}=0,05\left(mol\right)\)

\(n_{O_2}=\dfrac{0,9916}{24,79}=0,04\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,05}{2}< \dfrac{0,04}{1}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{Cu}=0,025\left(mol\right)\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,04-0,025=0,015\left(mol\right)\)

\(\Rightarrow V_{O_2\left(dư\right)}=0,015.24,79=0,37185\left(l\right)\)

c, \(n_{CuO}=n_{Cu}=0,05\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,05.80=4\left(g\right)\)

Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

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

_____0,4______0,6_______________0,6 (mol)

\(\Rightarrow a=C_{M_{H_2SO_4}}=\dfrac{0,6}{0,5}=1,2\left(M\right)\)

\(V=V_{H_2}=0,6.24,79=14,874\left(l\right)\)