Đáp án C 

nBr2=8160=0,05mol

⇒nC2H4=0,05mol

⇒mC2H4=0,05x28=1,4g

Chọn C

\(n_{Br_2}=\dfrac{8}{160}=0,05mol\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

0,05       0,05                        ( mol )

\(m_{C_2H_4}=0,05.28=1,4g\)

=> Chọn C

C

\(n_{CH_3COOC_2H_5}=\dfrac{4,4}{88}=0,05\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO₄(đ),t^o--> CH₃COOC₂H₅ + H₂O

               0,05<--------------------------------------0,05

=> \(m_{CH_3COOH\left(lý.thuyết\right)}=0,05.60=3\left(g\right)\)

=> \(m_{CH_3COOH\left(tt\right)}=\dfrac{3.100}{60}=5\left(g\right)\)

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

b, Phần này đề hỏi tính khối lượng gì bạn nhỉ?

c, \(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{Zn}=1\left(mol\right)\)

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{1.60}{36\%}=\dfrac{500}{3}\left(g\right)\)

a) \(n_{Na_2CO_3}=\dfrac{26,5}{106}=0,25\left(mol\right)\)

PTHH: Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

_____0,25------>0,5--------->0,5---->0,25

V_CO2 = 0,25.22,4 = 5,6(l)

b) m_HCl = 0,5.36,5 = 18,25(g)

=> a = \(\dfrac{18,25.100}{20}=91,25\left(g\right)\)

c) mNaCl = 0,5.58,5 = 29,25 (g)

d) PTHH: NaOH + HCl --> NaCl + H₂O

_________0,4---->0,4

=> Không trung hòa đc hết lượng axit

CH3COOH + C2H5OH => (H2SO4đ, t^o) pứ hai chiều> CH3COOC2H5 + H2O

nCH3COOH = m/M = 30/60 = 0.5 (mol)

nC2H5OH = m/M = 50/46 (mol)

Lập tỉ số: 50/46 > 0.5 => C2H5OH dư, CH3COOH hết

Theo phương trình ==> nCH3COOC2H5 = 0.5 (mol)

mCH3COOC2H5 theo phương trình = n.M = 0.5 x 88 = 44 (g)

mCH3COOC2H5 thu được = 44 x H% = 44x62.5/100 = 27.5 (g)

a, \(n_{CO_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\)

PT: \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

Theo PT: \(n_{CH_3COOH}=2n_{CO_2}=0,05\left(mol\right)\)

\(\Rightarrow C\%_{CH_3COOH}=\dfrac{0,05.60}{100}.100\%=3\%\)

b, Theo PT: \(n_{Na_2CO_3}=n_{CO_2}=0,025\left(mol\right)\Rightarrow m_{Na_2CO_3}=0,025.106=2,65\left(g\right)\)

\(n_{CH_3COONa}=2n_{CO_2}=0,05\left(mol\right)\Rightarrow m_{CH_3COONa}=0,05.82=4,1\left(g\right)\)

c, \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)

Theo PT: \(n_{C_2H_5OH\left(LT\right)}=n_{CH_3COOH}=0,05\left(mol\right)\)

Mà: H = 80%

\(\Rightarrow n_{C_2H_5OH\left(TT\right)}=\dfrac{0,05}{80\%}=0,0625\left(mol\right)\)

\(\Rightarrow m_{C_2H_5OH\left(TT\right)}=0,0625.46=2,875\left(g\right)\)

\(\Rightarrow V_{C_2H_5OH}=\dfrac{2,875}{0,8}=3,59375\left(ml\right)\)

\(\Rightarrow V_{C_2H_5OH\left(10^o\right)}=\dfrac{3,59375}{10}.100=35,9375\left(ml\right)\)

Chọn D

(m + 21)h