Đáp án:
\( {m_{Fe{\text{ dư}}}} = 2,8{\text{ gam}}\)
\( {V_{{H_2}}} = 3,36{\text{ lít}}\)
\( {{\text{m}}_{PbO}} = 33,45{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
Ta có:
\({n_{Fe}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = \frac{{14,7}}{{98}} = 0,15{\text{ mol}}\)
Vì \({n_{Fe}} > {n_{{H_2}S{O_4}}}\) nên \(Fe\) dư.
\({n_{Fe{\text{ dư}}}} = {n_{Fe}} - {n_{{H_2}S{O_4}}} = 0,2 - 0,15 = 0,05{\text{ mol}}\)
\( \to {m_{Fe{\text{ dư}}}} = 0,05.56 = 2,8{\text{ gam}}\)
Ta có:
\({n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,15{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)
\(PbO + {H_2}\xrightarrow{{{t^o}}}Pb + {H_2}O\)
Ta có:
\({n_{PbO}} = {n_{{H_2}}} = 0,15{\text{ mol}} \to {{\text{m}}_{PbO}} = 0,15.(207 + 16) = 33,45{\text{ gam}}\)
