Đáp án:
\({m_1} = 5,6{\text{ gam}}\)
\({m_2} = 7,2{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Fe + S\xrightarrow{{{t^o}}}FeS\)
\(FeS + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}S\)
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
Ta có:
\({n_{{H_2}}} + {n_{{H_2}S}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}}\)
\({n_S} = \frac{{1,6}}{{32}} = 0,05{\text{ mol = }}{{\text{n}}_{{H_2}S}} = {n_{{H_2}}}\)
\( \to {n_{Fe}} = {n_{FeS}} = 0,05{\text{ mol}}\)
\( \to {m_2} = {m_{Fe}} + {m_{FeS}} = 0,05.56 + 0,05.88 = 7,2{\text{ gam}}\)
\({n_{Fe{\text{ bđ}}}} = {n_{Fe}} + {n_{FeS}} = 0,1{\text{ mol}}\)
\( \to {m_1} = 0,1.56 = 5,6{\text{ gam}}\)
