Đáp án:
a) \({{\text{m}}_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = 63{\text{ gam}}\)
b) \(C{\% _{FeS{O_4}}} = 16,9\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
Ta có:
\({n_{Fe}} = {n_{{H_2}S{O_4}}} = {n_{FeS{O_4}}} = {n_{{H_2}}} = \frac{{7,2}}{{56}} = \frac{9}{{70}}{\text{ mol}}\)
\({m_{{H_2}S{O_4}}} = \frac{9}{{70}}.98 = 12,6{\text{ gam}} \to {{\text{m}}_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{12,6}}{{20\% }} = 63{\text{ gam}}\)
BTKL:
\({m_{dd\;{\text{sau phản ứng}}}} = {m_{Fe}} + {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} - {m_{{H_2}}} = 7,2 + 63 - \frac{9}{{70}}.2 = 69,94{\text{ gam}}\)
\({m_{FeS{O_4}}} = \frac{7}{{90}}.(56 + 96) = 11,82{\text{ gam}} \to C{\% _{FeS{O_4}}} = \frac{{11,82}}{{69,94}} = 16,9\% \)
