Đáp án:
\( {m_{{H_2}S{O_4}}} = 117,6{\text{ gam}}\)
\( {V_{{H_2}}} = 1,12{\text{ lít}}\)
\( {m_{FeS{O_4}}} = 7,6{\text{ gam}}\)
\({m_{{H_2}S{O_4}{\text{ dư}}}} = 112,7{\text{ gam}}\)
Giải thích các bước giải:
Ta có:
\({n_{{H_2}S{O_4}}} = {V_{{H_2}S{O_4}}}{C_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = 0,4.3 = 1,2{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 1,2.98 = 117,6{\text{ gam}}\)
Cho \(A\) tác dụng với \(Fe\)
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
Ta có:
\({n_{Fe}} = \frac{{2,8}}{{56}} = 0,05{\text{ mol < }}{{\text{n}}_{{H_2}S{O_4}}}\)
\( \to {n_{{H_2}}} = {n_{Fe}} = 0,05{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,05.22,4 = 1,12{\text{ lít}}\)
\({n_{FeS{O_4}}} = {n_{Fe}} = 0,05{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}{\text{ dư}}}} = 1,2 - 0,05 = 1,15{\text{ mol}}\)
\( \to {m_{FeS{O_4}}} = 0,05.(56 + 96) = 7,6{\text{ gam}}\)
\({m_{{H_2}S{O_4}{\text{ dư}}}} = 1,15.98 = 112,7{\text{ gam}}\)
