Đáp án:

\( {m_{CaC{O_3}}}  = 15{\text{ gam}}; {{\text{m}}_{FeO}} = 2,2{\text{ gam}}\)

\( \to m = 45,13877{\text{ gam}}\)

\({m_{CaC{O_3}}} = 15{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(CaC{O_3} + {H_2}S{O_4}\xrightarrow{{}}CaS{O_4} + C{O_2} + {H_2}O\)

\(FeO + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}O\)

Ta có:

\({n_{C{O_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol = }}{{\text{n}}_{CaC{O_3}}}\)

\( \to {m_{CaC{O_3}}} = 0,15.100 = 15{\text{ gam}} \to {{\text{m}}_{FeO}} = 17,2 - 15 = 2,2{\text{ gam}}\)

\({n_{FeO}} = \frac{{2,2}}{{72}} = \frac{{11}}{{360}}\)

\( \to {n_{{H_2}S{O_4}}} = 0,15 + \frac{{11}}{{360}} = \frac{{13}}{{72}}\)

\( \to {n_{{H_2}S{O_4}{\text{ tham gia}}}} = \frac{{13}}{{72.80\% }} = \frac{{65}}{{288}}\)

\( \to {m_{{H_2}S{O_4}}} = \frac{{65}}{{288}}.98 = 22,118{\text{ gam}}\)

\( \to m = \frac{{22,118}}{{49\% }} = 45,13877{\text{ gam}}\)

\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)

\({n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,15{\text{ mol}}\)

\( \to {m_{CaC{O_3}}} = 0,15.100 = 15{\text{ gam}}\)