1)
Oxit sắt có dạng là \(Fe_xO_y\).
Phản ứng xảy ra:
\(F{e_x}{O_y} + y{H_2}\xrightarrow{{{t^o}}}xFe + y{H_2}O\)
Ta có:
\({n_{Fe}} = \frac{{5,6}}{{56}} = 0,1{\text{ mol}} \to {{\text{n}}_{F{e_x}{O_y}}} = \frac{{{n_{Fe}}}}{x} = \frac{{0,1}}{x}\)
\( \to {M_{F{e_x}{O_y}}} = 56x + 16y = \frac{{7,2}}{{\frac{{0,1}}{y}}} = 72y \to 56x = 56y \to x:y = 1:1\)
Vậy oxit là \(FeO\).
2)
Phản ứng:
\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH\)
BTKL:
\({m_{dd{\text{ A}}}} = {m_{N{a_2}O}} + {m_{{H_2}O}} = 6,2 + 193,8 = 200{\text{ gam}}\)
\({n_{N{a_2}O}} = \frac{{6,2}}{{23.2 + 16}} = 0,1{\text{ mol}}\)
\( \to {n_{NaOH}} = 2{n_{N{a_2}O}} = 0,1.2 = 0,2{\text{ mol}}\)
\( \to {m_{NaOH}} = 0,2.40 = 8{\text{ gam}}\)
\( \to C{\% _{NaOH}} = \frac{8}{{200}} = 4\% \)
