1)
Ta có:
\({n_{Ca}} = \frac{{0,4}}{{40}} = 0,01{\text{ mol = }}{{\text{n}}_{CaO}}\)
\( \to {m_{CaO}} = 0,01.(40 + 16) = 0,56{\text{ gam}}\)
2)
\(2Zn + {O_2}\xrightarrow{{{t^o}}}2ZnO\)
\( \to {n_{Zn}} = \frac{{13}}{{65}} = 0,2{\text{ mol}} \to {{\text{n}}_{{O_2}}} = \frac{1}{2}{n_{Zn}} = 0,1{\text{ mol}}\)
\( \to {V_{{O_2}}} = 0,1.22,4 = 2,24{\text{ lít}} \to {{\text{V}}_{kk}} = 5{V_{{O_2}}} = 2,24.5 = 11,2{\text{ lít}}\)
