Đáp án:
$n_{HCl}=0,6\,\, mol$
\( m = 13,05{\text{ gam}}\)
Giải thích các bước giải:
Phương trình hóa học:
\(Mn{O_2} + 4HCl\xrightarrow{{}}MnC{l_2} + C{l_2} + 2{H_2}O\)
\(2F{\text{e}} + 3C{l_2}\xrightarrow{{{t^o}}}2F{\text{e}}C{l_3}\)
Ta có:
\({n_{F{\text{e}}C{l_3}}} = \dfrac{{16,25}}{{56 + 35,5.3}} = 0,1{\text{ mol}} \\\to {{\text{n}}_{C{l_2}}} = \dfrac{3}{2}{n_{F{\text{e}}C{l_3}}} = 0,15{\text{ mol}}\)
\( \to {n_{Mn{O_2}}} = {n_{C{l_2}}} = 0,15{\text{ mol;}}{{\text{n}}_{HCl}} = 4{n_{C{l_2}}} = 0,6{\text{ mol}}\)
\( \to m = {m_{Mn{O_2}}} = 0,15.(55 + 16.2) = 13,05{\text{ gam}}\)