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)

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

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

a, Chất rắn là Ag.

⇒ m_Ag = 10,8 (g)

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

Ta có: \(n_{H_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\)

⇒ m_Al = 0,4.27 = 10,8 (g)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{10,8}{10,8+10,8}.100\%=50\%\\\%m_{Ag}=50\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=1,2\left(mol\right)\)

\(\Rightarrow V=\dfrac{1,2}{1,5}=0,8\left(l\right)=800\left(ml\right)\)

\(n_{AlCl_3}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\Rightarrow C_{M_{AlCl_3}}=\dfrac{0,4}{0,8}=0,5\left(M\right)\)

a, \(2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\)

b, \(n_{Al}=\dfrac{12,15}{27}=0,45\left(mol\right)\)

\(n_{Cl_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,45}{2}>\dfrac{0,6}{3}\), ta được Al dư.

Theo PT: \(n_{Al\left(pư\right)}=n_{AlCl_3}=\dfrac{2}{3}n_{Cl_2}=0,4\left(mol\right)\)

⇒ n_{Al (dư)} = 0,45 - 0,4 = 0,05 (mol)

⇒ m_{Al (dư)} = 0,05.27 = 1,35 (g)

b, m_AlCl3 = 0,4.133,5 = 53,4 (g)

Hỗn hợp khí thu được gồm: CO và CO₂

Ta có: \(\left\{{}\begin{matrix}n_{CO}+n_{CO_2}=\dfrac{4,958}{24,79}=0,2\\28n_{CO}+44n_{CO_2}=19.2.0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{CO}=0,075\left(mol\right)\\n_{CO_2}=0,125\left(mol\right)\end{matrix}\right.\)

PT: \(C+\dfrac{1}{2}O_2\underrightarrow{t^o}CO\)

\(C+O_2\underrightarrow{t^o}CO_2\)

Theo PT: \(n_C=n_{CO}+n_{CO_2}=0,2\left(mol\right)\)

⇒ m_C = 0,2.12 = 2,4 (g)