PT: \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
Ta có: \(n_{Cl_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{KMnO_4\left(LT\right)}=\dfrac{2}{5}n_{Cl_2}=0,2\left(mol\right)\\n_{HCl\left(LT\right)}=\dfrac{16}{5}n_{Cl_2}=1,6\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{KMnO_4\left(LT\right)}=0,2.158=31,6\left(g\right)\\V_{ddHCl\left(LT\right)}=\dfrac{1,6}{2}=0,8\left(l\right)\end{matrix}\right.\)
Mà: H% = 75%
\(\Rightarrow\left\{{}\begin{matrix}m_{KMnO_4\left(TT\right)}=\dfrac{31,6}{75\%}\approx42,13\left(g\right)\\V_{ddHCl\left(TT\right)}=\dfrac{0,8}{75\%}\approx1,067\left(l\right)\end{matrix}\right.\)
Bạn tham khảo nhé!
