Đáp án:
\({{\text{V}}_{{H_2}}} = 4,48{\text{ lít}}\)
\({{\text{m}}_{{H_2}S{O_4}}} = 19,6{\text{ gam}}\)
\({\text{\% }}{{\text{m}}_{FeS{O_4}}} = 16,43\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
Ta có:
\({n_{{H_2}}} = {n_{Fe}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
\({n_{{H_2}S{O_4}}} = {n_{Fe}} = 0,2{\text{ mol}} \to {{\text{m}}_{{H_2}S{O_4}}} = 0,2.98 = 19,6{\text{ gam}}\)
\({m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{19,6}}{{11,25\% }} = 174,22{\text{ gam}}\)
BTKL: \({m_{dd{\text{ sau phản ứng}}}} = 11,2 + 174,22 - 0,2.2 = 185,02{\text{ gam}}\)
\({n_{FeS{O_4}}} = {n_{Fe}} = 0,2{\text{ mol}} \to {{\text{m}}_{FeS{O_4}}} = 0,2.152 = 30,4{\text{ gam}} \to {\text{\% }}{{\text{m}}_{FeS{O_4}}} = \frac{{30,4}}{{185,02}} = 16,43\% \)
