\(n_{H_2} = \dfrac{5,78125-5,625}{2} = 0,078125(mol)\)

Gọi CTPT của anđehit cần tìm : C_nH_2n+1CHO

\(C_nH_{2n+1}CHO + H_2 \xrightarrow{t^o,H_2} C_nH_{2n+1}CH_2OH\)

Suy ra :

\(M_{anđehit} = 14n + 30 = \dfrac{5,625}{0,078125} = 72\\ \Rightarrow n = 3\)

Vậy CTPT của anđehit là C₄H₈O

\(C_6H_5OH + NaOH \to C_6H_5ONa + H_2O\\ n_{C_6H_5OH}= n_{NaOH} = 0,4.0,3 = 0,12(mol)\\ 2C_6H_5OH + 2Na \to 2C_6H_5ONa +H_2\\ 2C_2H_5OH + 2Na \to 2C_2H_5ONa + H_2\\ n_{H_2} =\dfrac{1}{2}n_{C_6H_5OH} + \dfrac{1}{2}n_{C_2H_5OH} = 0,06 + \dfrac{1}{2}n_{C_2H_5OH} = \dfrac{3,584}{22,4} = 0,16(mol)\\ \Rightarrow n_{C_2H_5OH} = 0,2\\ \Rightarrow m_A = 0,12.94 + 0,2.46 = 20,48(gam) \)

\(X : C_nH_{2n+1}OH\\ 2C_nH_{2n+1}OH + 2Na \to 2C_nH_{2n+1}ONa + H_2\\ n_{H_2} = \dfrac{0,672}{22,4} = 0,03(mol)\\ n_X = 2n_{H_2} = 0,06(mol)\\ \Rightarrow M_X = 14n + 18 = \dfrac{2,76}{0,06} =46\\ \Rightarrow n = 2\)

Vậy CTPT của X là C₂H₆O

Câu 6 :

\(2KClO_3 \xrightarrow{t^o} 2KCl +3 O_2\\ a) n_{O_2} = \dfrac{36}{32} = 1,125(mol)\\ n_{KClO_3}= \dfrac{2}{3}n_{O_2} =0,75(mol)\\ m_{KClO_3} = 0,75.122,5 = 91,875(gam)\\ b) n_{O_2} = \dfrac{33,6}{22,4} = 1,5(mol)\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 1(mol)\\ m_{KClO_3} = 1.122,5 = 122,5(gam)\)

Câu 4 :

\(a) 2KClO_3 \xrightarrow{t^o,MnO_2} 2KCl + 3O_2\\ b) n_{O_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 0,1(mol)\\ m_{KClO_3} = 0,1.122,5 = 12,25(gam)\\ c) CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ V_{CH_4} = \dfrac{1}{2}V_{O_2} = 1,68(lít)\)

\(a) n_{CO_2} = \dfrac{7,84}{22,4} = 0,35(mol)\ ; n_{H_2O} = \dfrac{8,1}{18} = 0,45(mol)\\ n_{CO_2} < n_{H_2O} \Rightarrow \text{Ancol no, đơn chức}\\ n_{Ancol} = n_{H_2O} - n_{CO_2} = 0,1(mol)\)

Số nguyên tử C trung bình = \(\dfrac{0,35}{0,1 } = 3,5\)

Vậy hai ancol là : \(C_3H_7OH\ ; C_4H_{9}OH\)

\(b) n_{C_3H_7OH} = a(mol) ; n_{C_4H_9OH} = b(mol)\\ n_{ancol} = a+ b = 0,1(mol)\\ n_{CO_2} = 3a + 4b = 0,35(mol)\\ \Rightarrow a = b = 0,05\\ \%m_{C_3H_7OH} = \dfrac{0,05.60}{0,05.60 + 0,05.74}.100\% = 44,78\%\\ \%m_{C_4H_9OH} = 100\% -44,78\% = 55,22\%\)

\(c)2 C_3H_7OH + 2Na \to 2C_3H_7ONa + H_2\\ 2C_4H_9OH + 2Na \to 2C_4H_9ONa + H_2\\ n_{H_2} = \dfrac{1}{2}n_{ancol} = 0,05(mol)\\ V_{H_2} = 0,05.22,4 = 1,12(lít)\)

Oxit tác dụng với nước : \(CaO,Na_2O,SO_3,P_2O_5,K_2O,SO_2,N_2O_5,CO_2,BaO\)

Phương trình hóa học :

\(CaO + H_2O \to Ca(OH)_2\\ Na_2O + H_2O \to 2NaOH\\ SO_3 + H_2O \to H_2SO_4\\ P_2O_5 + 3H_2O \to 2H_3PO_4\\ K_2O + H_2O \to 2KOH\\ SO_2 + H_2O \to H_2SO_3\\ N_2O_5 + H_2O \to 2HNO_3\\ CO_2 + H_2O \leftrightharpoons H_2CO_3\\ BaO + H_2O \to Ba(OH)_2\)

\(2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ P_2O_5 + 3H_2O \to 2H_3PO_4\\ 3Na + H_3PO_4 \to Na_3PO_4 + \dfrac{3}{2}H_2\)

Theo PTHH :

\( n_{H_2} = n_{H_2O} = \dfrac{3,6}{18} = 0,2(mol)\\ \Rightarrow V_{H_2} = 0,2.22,4 = 4,48(lít)\)