\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2mol\\ 2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\\ n_{HCl}=\dfrac{0,2}{2}\cdot16=1,6mol\\ V_{HCl}=\dfrac{1,6}{0,5}=3,2l\)
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