1. \(N_2+3H_2\underrightarrow{t^o}2NH_3\)

2. \(4Na+O_2\underrightarrow{t^o}2Na_2O\)

3. \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

4. \(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

5. \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

6. \(2HCl+Zn\rightarrow ZnCl_2+H_2\)

Có lẽ đề hỏi khối lượng các KL trong hh đầu chứ bạn nhỉ?

Ta có: 27n_Al + 56n_Fe + 64n_Cu = 35,8 (1)

BTNT Al, Fe: n_AlCl3 = n_Al, n_FeCl2 = n_Fe

⇒ 133,5n_Al + 127n_Fe = 52,1 (1)

\(n_{H_2}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)

BT e, có: 3n_Al + 2n_Fe = 0,5.2 (3)

Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Fe}=0,2\left(mol\right)\\n_{Cu}=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Cu}=0,3.64=19,2\left(g\right)\end{matrix}\right.\)

X là hidrocarbon nên X chỉ chứa C và H nên m_X = m_C + m_H bạn nhé.

Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

___0,2______0,2_______0,2____0,2 (mol)

b, V_H2 = 0,2.24,79 = 4,958 (l)

c, \(m_{ddH_2SO_4}=\dfrac{0,2.98}{10\%}=196\left(g\right)\)

d, m _{dd sau pư} = 13 + 196 - 0,2.2 = 208,6 (g)

\(\Rightarrow C\%_{ZnSO_4}=\dfrac{0,2.161}{208,6}.100\%\approx15,4\%\)

e, \(H_2SO_4+Ca\left(OH\right)_2\rightarrow CaSO_4+2H_2O\)

______0,2_______0,2 (mol)

\(m_{ddCa\left(OH\right)_2}=\dfrac{0,2.74}{20\%}=74\left(g\right)\)

\(\Rightarrow V_{Ca\left(OH\right)_2}=\dfrac{74}{1,12}\approx66,1\left(ml\right)\)

a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

b, Ta có: \(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,1\left(mol\right)\)

\(n_{CO_2}=\dfrac{11}{44}=0,25\left(mol\right)\)

Theo PT: \(n_{CH_4}=n_{CO_2}=0,25\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\%n_{CH_4}=\dfrac{0,25}{0,25+0,1}.100\%\approx71,4\%\\\%V_{C_2H_4}\approx28,6\%\end{matrix}\right.\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

BTNT H: n_HCl = 2n_H2 = 0,4 (mol)

\(\Rightarrow C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

Gọi: n_CH4 = n_C2H6 = a (mol)

PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

_____a____________a (mol)

\(2C_2H_6+7O_2\underrightarrow{t^o}4CO_2+6H_2O\)

___a____________2a (mol)

⇒ (a + 2a).44 = 60

⇒ a = 5/11 (mol)

⇒ m = 5/11.16 + 5/11.30 = 230/11 (g)

\(n_{CH_4}=\dfrac{3,2}{16}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,1}{2}\), ta được CH₄ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{1}{2}n_{O_2}=0,05\left(mol\right)\\n_{H_2O}=n_{O_2}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ m = 0,05.44 + 0,1.18 = 4 (g)

a, \(n_C=n_{CO_2}=n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

⇒ V = 0,1.24,79 = 2,479 (l)

Có: m_X = m_C + m_H ⇒ m_H = 1,6 - 0,1.12 = 0,4 (g) ⇒ n_H = 0,4 (mol)

⇒ n_H2O = 0,2 (mol)

⇒ m = 0,2.18 = 3,6 (g)

b, Có: n_H2O > n_CO2 → X là alkan.

Gọi CTPT của X là C_nH_2n+2

\(\Rightarrow\dfrac{n}{2n+2}=\dfrac{0,1}{0,4}\Rightarrow n=1\)

→ X là CH₄. Tên gọi: Methane.

c, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

__________0,2____0,1 (mol)

⇒ V_O2 = 0,2.24,79 = 4,958 (l)

a, \(n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\)

PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

______0,1____0,2___0,1_____0,2 (mol)

m = 0,1.16 = 1,6 (g)

V = 0,1.24,79 = 2,479 (l)

b, V_{O2 (đkc)} = 0,2.24,79 = 4,958 (l)

V_{O2 (đktc)} = 0,2.22,4 = 4,48 (l)