Đáp án:
a) PTHH: $CaC{O_3}{\text{ }}\xrightarrow{{{t^o}}}CaO{\text{ }} + {\text{ }}C{O_2}$
b) $m_{CaCO_3} = 25 (g)$
c) ${m_{CaO}}{\text{ }} = {\text{ }}14{\text{ }}\,\,\left( g \right)$
Giải thích các bước giải:
${n_{C{O_2}{\text{ }}}} = {\text{ }}\dfrac{{5,6}}{{22,4}} = {\text{ }}0,25{\text{ }}\,\,mol$
a) PTHH: $CaC{O_3}{\text{ }}\xrightarrow{{{t^o}}}CaO{\text{ }} + {\text{ }}C{O_2}$
b) Theo PTHH: ${n_{CaC{O_{3{\text{ }}}}}} = {\text{ }}{n_{C{O_2}{\text{ }}}} = {\text{ }}0,25{\text{ }}mol\;$
=> ${m_{CaC{O_3}{\text{ }}}} = {\text{ }}0,25.100{\text{ }} = {\text{ }}25{\text{ }}\left( g \right)$
c) Theo PTHH: ${n_{CaO}}{\text{ }} = {\text{ }}{n_{C{O_2}{\text{ }}}} = {\text{ }}0,25{\text{ }}mol$
=> ${m_{CaO}}{\text{ }} = {\text{ }}0,25.56{\text{ }} = {\text{ }}14{\text{ }}\left( g \right)$
