\(n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

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

Theo PT: \(n_{C_2H_4}=\dfrac{1}{3}n_{O_2}=0,5\left(mol\right)\)

\(\Rightarrow V_{C_2H_4}=0,5.22,4=11,2\left(l\right)\)

Đáp án: C

C.11,2l

\(n_{CH_3COOH}=\dfrac{30}{60}=0,5\left(mol\right)\)

\(n_{C_2H_5OH}=\dfrac{50}{46}=\dfrac{25}{23}\left(mol\right)\)

PT: \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H₂SO₄ đặc, t^o)

Xét tỉ lệ: \(\dfrac{0,5}{1}< \dfrac{\dfrac{25}{23}}{1}\), ta được C₂H₅OH dư.

Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,5\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=0,5.88=44\left(g\right)\) < m_{CH3COOC2H5 (TT)}

Bạn xem lại đề nhé, lượng thu được ở lí thuyết phải lớn hơn thực tế.

Gọi: n_{OH (trong bazo)} = n_{Cl (trong muối)} = x (mol)

⇒ m_KL + 17x = 20,5 

và: m_KL + 35,5x = 29,75

⇒ 20,5 - 17x = 29,75 - 35,5x ⇒ x = 0,5 (mol)

BTNT Cl, có: n_HCl = n_{Cl (trong muối)} = 0,5 (mol)

\(\Rightarrow V_{HCl}=\dfrac{0,5}{1}=0,5\left(l\right)=500\left(ml\right)\)

cho mình hỏi là \(m_{KL}\) là gì ạ

\(n_{C_2H_2}=\dfrac{44,8}{22,4}=2\left(mol\right)\)

PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Theo PT: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=5\left(mol\right)\)

\(\Rightarrow V_{O_2}=5.22,4=112\left(l\right)\)

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

\(C_2H_5OH+K\rightarrow C_2H_5OK+\dfrac{1}{2}H_2\)

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

b, \(CH_3COOH+Na\rightarrow CH_3COONa+\dfrac{1}{2}H_2\)

\(CH_3COOH+K\rightarrow CH_3COOK+\dfrac{1}{2}H_2\)

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

\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

Giải giúp

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(m_{HCl}=100.14,6\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{Zn}=0,2\left(mol\right)\\n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)

Ta có: m _{dd sau pư} = 6,5 + 100 - 0,1.2 = 106,3 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{106,3}.100\%\approx12,79\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,2.36,5}{106,3}.100\%\approx6,87\%\end{matrix}\right.\)

Em cảm ơn idol

a, \(m_{HCl}=150.14,6\%=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Theo PT: \(n_{Zn}=n_{ZnCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,3.65=19,5\left(g\right)\)

b, \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

c, Ta có: m _{dd sau pư} = 19,5 + 150 - 0,3.2 = 168,9 (g)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,3.136}{168,9}.100\%\approx24,16\%\)

Em cảm ơn nhiều ạ :3! 

\(n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)\)

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

Theo PT: \(n_{KMnO_4}=2n_{O_2}=1\left(mol\right)\Rightarrow m_{KMnO_4}=1.158=158\left(g\right)\)

Ta có: \(\dfrac{m_{CO_2}}{m_{H_2O}}=\dfrac{77}{18}\Rightarrow\dfrac{n_{CO_2}}{n_{H_2O}}=\dfrac{7}{4}\Rightarrow\dfrac{n_C}{n_H}=\dfrac{7}{8}\)

Gọi CTPT của X là C_xH_y.

⇒ x:y = 7:8

→ CTPT của X có dạng là (C₇H₈)_n

Mà: \(n_{X\left(5,52\left(g\right)\right)}=\dfrac{1,68}{28}=0,06\left(mol\right)\Rightarrow M_X=\dfrac{5,52}{0,06}=92\left(g/mol\right)\)

\(\Rightarrow n=\dfrac{92}{12.7+1.8}=1\)

Vậy: X là C₇H₈.

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b, \(m_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)

Theo PT: \(n_{MgCl_2}=n_{Mg}=0,3\left(mol\right)\Rightarrow m_{MgCl_2}=0,3.95=28,5\left(g\right)\)

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

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\left(M\right)\)