Đáp án:
18 A
20 A
Giải thích các bước giải:
\(\begin{array}{l}
18)\\
N{a_2}C{O_3} + BaC{l_2} \to BaC{O_3} + 2NaCl\\
{K_2}C{O_3} + BaC{l_2} \to BaC{O_3} + 2KCl\\
{n_{BaC{O_3}}} = \dfrac{{3,94}}{{197}} = 0,02\,mol\\
{n_{BaC{l_2}}} = {n_{BaC{O_3}}} = 0,02\,mol\\
BTKL\\
{m_{hh}} + {m_{BaC{l_2}}} = m + {m_{BaC{O_3}}}\\
\Rightarrow m = 2,44 + 0,02 \times 208 - 3,94 = 2,66g\\
20)\\
F{e_2}{O_3} + 3CO \xrightarrow{t^0} 3C{O_2} + 2Fe\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{F{e_2}{O_3}}} = \dfrac{{32}}{{160}} = 0,2\,mol\\
{n_{C{O_2}}} = 3{n_{F{e_2}{O_3}}} = 0,6\,mol\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,6\,mol\\
a = {m_{CaC{O_3}}} = 0,6 \times 100 = 60g
\end{array}\)
