m b e n z e n = 11,7 85 % = 13,765 %

Đáp án B

n_C6H5Br = 0,2 mol

C₆H₆ + Br₂ → C₆H₅Br + HBr

0,2 <-----------0,2

⇒ m_C6H6 = 0,2.78.\(\dfrac{100}{85}\) = 18,35 (g)

C6H6 + Br2 \(\rightarrow\) C6H5Br + HBr

n C6H5Br = 15,7 : 157 = 0,1 mol

Theo PTHH : n C6H6 = n C6H5Br = 0,1 mol

\(\Rightarrow\) m C6H6 = 0,1 . 78 = 7,8 g

Vì hiệu suất phản ứng đạt 80% nên :

m C6H6 cần dùng = 7,8 . 100 : 80 = 9,75 g

\(C_6H_6+Br_2\underrightarrow{^{xt}}C_6H_5Br+HBr\)

Ta có: \(n_{C6H5Br}=\frac{39,25}{12.6+5+80}=0,25\left(mol\right)\)

\(\rightarrow n_{C6H6_{pu}}=n_{C6H5Br}=0,25\left(mol\right)\)

\(\rightarrow n_{C6H6_{tham.gia}}=\frac{0,25}{85\%}=\frac{5}{17}\left(mol\right)\)

\(\Rightarrow m_{C6H6_{tham.gia}}=\frac{5}{17}.12.6+6=22,94\left(g\right)\)

\(a,\%m_{Cl}=\dfrac{213}{291}.100\%=73,2\%\\ b,n_{C_6H_6Cl_6}=\dfrac{58,2}{291}=0,2\left(mol\right)\)

PTHH: \(C_6H_6+3Cl_2\xrightarrow[]{askt}C_6H_6Cl_6\)

             0,2<--------------------0,2

\(\rightarrow m_{C_6H_6}=\dfrac{0,2.78}{40\%}=39\left(g\right)\)

\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: 2KMnO₄ --t^o--> K₂MnO₄ + MnO₂ + O₂

               1<------------------------------------0,5

=> \(m_{KMnO_4\left(pthh\right)}=1.158=158\left(g\right)\)

=> \(m_{KMnO_4\left(tt\right)}=\dfrac{158.100}{80}=197,5\left(g\right)\)