a)
Ta có: \({n_{CaC{O_3}}} = 0,5{\text{ mol}}\)
\( \to {n_{Ca}} = {n_C} = {n_{CaC{O_3}}} = 0,5{\text{ mol}}\)
\( \to {n_O} = 3{n_{CaC{O_3}}} = 3.0,5 = 1,5{\text{ mol}}\)
b)
Ta có:
\({n_{CaC{O_3}}} = \frac{{20}}{{40 + 12 + 16.3}} = 0,2{\text{ mol}}\)
\( \to {n_{Ca}} = {n_C} = {n_{CaC{O_3}}} = 0,2{\text{ mol}}\)
\({n_O} = 3{n_{CaC{O_3}}} = 0,2.3 = 0,6{\text{ mol}}\)
\( \to {m_{Ca}} = 0,2.40 = 8{\text{ gam}}\)
\({m_C} = 0,2.12 = 2,4{\text{ gam}}\)
\({m_O} = 0,6.16 = 9,6{\text{ gam}}\)
