Ta có: \(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)

PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

_____0,05____0,05 (mol)

\(\Rightarrow V_{C_2H_4}=0,05.22,4=1,12\left(l\right)\)

\(V_{CH_4}=5,6-1,12=4,48\left(l\right)\)

Ta có: n_OH^- = 2n_H2 = 0,4 (mol) = n_H⁺

Gọi: V _{dd axit} = a (l)

⇒ n_H⁺ = a + 0,5.2.a = 0,4

⇒ a = 0,2 (l) = 200 (ml)

→ Đáp án: C

Đáp án: C

Ta có: \(n_{C_2H_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

____0,25___0,75 (mol)

\(\Rightarrow V_{O_2}=0,75.22,4=16,8\left(g\right)\)

\(V_{kk}=\dfrac{V_{O_2}}{20\%}=84\left(l\right)\)

Ta có: n_N2 + n_H2 = 1,3 (mol) (1)

Mà: d_hh/O2 = 0,3125

\(\Rightarrow\dfrac{28n_{N_2}+2n_{H_2}}{1,3}=0,3125.32\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{N_2}=0,4\left(mol\right)\\n_{H_2}=0,9\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{N_2}=\%n_{N_2}=\dfrac{0,4}{1,3}.100\%\approx30,77\%\\\%V_{H_2}\approx69,23\%\end{matrix}\right.\)

Trong 1 phần: 24n_Na + 27n_Al + 56n_Fe = 56,7:3 (1)

- Phần 1: 2n_H2 = n_Na + 3n_Na 

⇒ 4n_Na = 0,2.2 (2)

- Phần 2: 

BT e: n_Na + 3n_Al = 2n_H2 = 0,35.2 (3)

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

BT e: 2n_H2 = n_Na + 3n_Al + 2n_Fe 

⇒ n_H2 = 0,55 (mol)

⇒ V_H2 = 0,55.22,4 = 12,32 (l)

→ Đáp án: B

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PT: \(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)

________0,2______0,2________0,2_____0,1 (mol)

a, m_C2H5OH = 0,2.46 = 9,2 (g)

b, m_Na = 0,2.23 = 4,6 (g)

m_C2H5ONa = 0,2.68 = 13,6 (g)

(1) \(C_2H_2+H_2\underrightarrow{^{Pd/PbCO_3,t^o}}C_2H_4\)

(2) \(C_2H_4+H_2O\underrightarrow{^{t^o,xt}}C_2H_5OH\)

(3) \(C_2H_5OH+O_2\xrightarrow[_{mengiam}]{^{t^o}}CH_3COOH+H_2O\)

(4) \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (Điều kiện: t^o, H₂SO₄ đặc)

(5) \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)

(6) \(2CH_3COOH+Ca\rightarrow\left(CH_3COO\right)_2Ca+H_2\)

Ta có: 102n_Al2O3 + 80n_CuO = 26,2 (1)

PT: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

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

Theo PT: \(n_{H_2SO_4}=3n_{Al_2O_3}+n_{CuO}=0,25.2=0,5\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Al_2O_3}=0,1\left(mol\right)\\n_{CuO}=0,2\left(mol\right)\end{matrix}\right.\)

⇒ m_Al2O3 = 0,1.102 = 10,2 (g)

m_CuO = 0,2.80 = 16 (g)

a, \(2CH_3COOH+CaCO_3\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\)

b, \(n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{CaCO_3}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,6}{0,25}=2,4\left(M\right)\)

c, \(n_{\left(CH_3COO\right)_2Ca}=n_{CaCO_3}=0,3\left(mol\right)\)

\(\Rightarrow m_{\left(CH_3COO\right)_2Ca}=0,3.158=47,4\left(g\right)\)