Đáp án C

Đáp án A

Ta có phản ứng: 

C2H4(OH)2 + 2K → C2H4(OK)2 + H2.

+ Tăng giảm khối lượng ta có 

nK pứ = 8,74 ÷ 38 = 0,23 mol.

⇒ mK = 8,97 gam

Đáp án A

Câu 1:

\(Mg+Br_2\rightarrow MgBr_2\\ n_{Br_2}=\dfrac{11,2}{160}=0,07\left(mol\right)=n_{Mg}=n_{MgBr_2}\\ a=m_{Mg}=0,07.24=1,68\left(g\right)\\ m_{MgBr_2}=184.0,07=12,88\left(g\right)\)

Mg+Br2->MgBr2

0,07--0,07----0,07

n Br2=\(\dfrac{11,2}{160}\)=0,07 mol

=>m Mg=0,07.24=1,68g

=>m MgBr2=0,07.184=12,88g

Câu 2:

\(n_{Al}=\dfrac{14,58}{27}=0,54\left(mol\right)\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ n_{AlCl_3}=n_{Al}=0,54\left(mol\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,54=0,81\left(mol\right)\\ V_{Cl_2\left(đktc\right)}=0,81.22,4=18,144\left(l\right)\\ m_{AlCl_3}=0,54.133,5=72,09\left(g\right)\)

Câu 2:

\(n_{MgBr_2}=\dfrac{14,72}{184}=0,08\left(mol\right)\\ Mg+Br_2\rightarrow MgBr_2\\ n_{Mg}=n_{Br_2}=n_{MgBr_2}=0,08\left(mol\right)\\ a=m_{Mg}=24.0,08=1,92\left(g\right)\\ m_{Br_2}=160.0,08=12,8\left(g\right)\)

Câu 1:

\(n_{AlBr_3}=\dfrac{106,8}{267}=0,4\left(mol\right)\\ 2Al+3Br_2\rightarrow2AlBr_3\\ n_{Al}=n_{AlBr_3}=0,4\left(mol\right)\\ n_{Br_2}=\dfrac{3}{2}.0,4=0,6\left(mol\right)\\ a=m_{Al}=0,4.27=10,8\left(g\right)\\ m_{Br_2}=160.0,6=96\left(g\right)\)

2Al+3Br₂-to>2AlBr₃

0,4-----0,6-------0,4 mol

n AlBr3=\(\dfrac{106,8}{267}\)=0,4 mol

=>a=mAl=0,4.27=10,8g

=>mBr₂=0,6.160=96g

Câu 2

Mg+Br₂->MgBr₂

0,08--0,08---0,08 mol

n _MgBr2=\(\dfrac{14,72}{184}\)=0,08 mol

=>a=mMg=0,08.24=1,92g

=>m _Br2=0,08.160=12,8g

Câu 1:

\(2Na+Br_2\rightarrow2NaBr\\ n_{NaBr}=\dfrac{61,8}{103}=0,6\left(mol\right)\\ n_{Na}=n_{NaBr}=0,6\left(mol\right)\\ n_{Br_2}=\dfrac{0,6}{2}=0,3\left(mol\right)\\ \Rightarrow a=m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Br_2}=0,3.160=48\left(g\right)\\ m_{ddBr_2}=\dfrac{48}{5\%}=960\left(g\right)\)

Câu 2:

\(2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{FeCl_3}=\dfrac{40,625}{162,5}=0,25\left(mol\right)\\ n_{Fe}=n_{FeCl_3}=0,25\left(mol\right)\\ \Rightarrow m=m_{Fe}=0,25.56=14\left(g\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,25=0,375\left(mol\right)\\ V_{Cl_2\left(đktc\right)}=0,375.22,4=8,4\left(l\right)\)

2Fe+3Cl₂-to>2FeCl₃

_{0,25------0,375------0,25 mol}

n _FeCl3=\(\dfrac{40,625}{162,5}\)=0,25 mol

=>m _Fe=0,25.56=14g

=>V _Cl2=0,375.22,4=8,4g

1)

a, \(n_{Al}=\dfrac{15,3}{102}=0,15\left(mol\right)\)

PTHH: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

Mol:     0,15         0,9        0,3      

\(m_{ddHCl}=\dfrac{0,9.36,5.100}{20}=164,25\left(g\right)\)

b, m_{dd sau pứ} = 15,3 + 164,25 = 179,55 (g)

c, \(C\%_{ddAlCl_3}=\dfrac{0,3.133,5.100\%}{179,55}=22,31\%\)

2)

a, \(m_{HCl}=54,75.20\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: Al₂O₃  + 6HCl → 2AlCl₃ + 3H₂O

Mol:      0,05        0,3          0,1

\(m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)

b, m_{dd sau pứ} = 5,1 + 54,75 = 59,85 (g) 

\(C\%_{ddAlCl_3}=\dfrac{0,1.133,5.100\%}{59,85}=22,31\%\)

Chọn đáp án A

nNaOH = 0,3 Þ nC₆H₅OH + nCH₃COOH + nCH₃COOC₂H₅ = 0,3

nH₂ = 0,11 Þ nC₆H₅OH + nCH₃COOH = 0,22 Þ nCH₃COOC₂H₅ = 0,08 = nC₂H₅OH

BTKL Þ mX + 0,3x40 = mY + 0,22x18 + 0,08x46 Þ mY - mX = 4,36.

Đáp án C