Đặt \(n_{KClO_3}=c\left(mol\right)\)
\(n_{KCl}=0,2\left(mol\right)\)
\(BTNT.Cr\Rightarrow n_{K_2Cr_2O_7}=\frac{1}{2}n_{CrCl_3}=0,5b\left(mol\right)\\ BTNT.Mn\Rightarrow n_{KMnO_4}=n_{MnCl_2}=a\left(mol\right)\\ \Rightarrow m_X=158x+122,5c+147b=57,45\left(1\right)\\ BTNT.K\Rightarrow n_{KCl}=n_{KMnO_4}+n_{KClO_3}+2n_{K_2Cr_2O_7}\\ =a+b+c=0,4\left(2\right)\\ \text{Mà }2a=b\left(3\right)\)
\(\left(1\right);\left(2\right);\left(3\right)\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\\c=0,1\end{matrix}\right.\)
QT nhường e:
Mn+7 + 5e ----> Mn+2
0,1_____0,5
Cl+5 + 6e ----> Cl-
0,1____0,6
Cr2+6 + 6e ----> 2Cr+3
_______0,6______0,2
QT nhận e: 2Cl- ----> Cl2 + 2e
BT e \(\Rightarrow n_{Cl_2}=0,85\left(mol\right)\)
\(\Rightarrow V_{Cl_2}=19,04\left(l\right)\)
Chọn A
Gọi số mol KMnO4;KClO3;K2Cr2O7 lần lượt là x, y, z.
\(\Leftrightarrow158x+122,5y+294z=57,45\left(1\right)\)
\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(KClO_3+6HCl\rightarrow KCl+3Cl_2+3H_2O\)
\(K_2Cr_2O_7+14HCl\rightarrow2KCl+2CrCl_3+3Cl_2+7H_2O\)
\(n_{KCl}=\frac{29,8}{74,5}=0,4\left(mol\right)\)
\(\Leftrightarrow x+y=2z=0,4\left(2\right)\)
\(\left\{{}\begin{matrix}n_{MgCl2}:x\left(mol\right)\\n_{CrCl3}:y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\frac{x}{2z}=\frac{1}{2}\Rightarrow x-z=0\left(3\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,1\left(mol\right)\\z=0,1\left(mol\right)\end{matrix}\right.\)
\(\Leftrightarrow n_{Cl2}=2,5x+3y+3z=2,5.0,1+3.0,1+3.0,1=0,85\left(mol\right)\)
\(\Rightarrow V=0,85.22,4=19,04\left(l\right)\)
Vậy đáp án đúng : A
