Đáp án:
\(\begin{array}{l}
a)\\
R:Natri(Na)\\
b)\\
\% {m_{N{a_2}S{O_4}}} = 57,26\% \\
\% {m_{N{a_2}C{O_3}}} = 42,74\%
\end{array}\]
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{P_1}:\\
{R_2}C{O_3} + {H_2}S{O_4} \to {R_2}S{O_4} + C{O_2} + {H_2}O\\
{P_2}:\\
{R_2}C{O_3} + BaC{l_2} \to 2RCl + BaC{O_3}\\
{R_2}S{O_4} + BaC{l_2} \to 2RCl + BaS{O_4}\\
{n_{C{O_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol \Rightarrow {n_{{R_2}C{O_3}}} = {n_{C{O_2}}} = 0,15\,mol\\
{n_{BaS{O_4}}} = \dfrac{{64,5 - 0,15 \times 197}}{{233}} = 0,15\,mol\\
\Rightarrow {n_{{R_2}S{O_4}}} = {n_{BaS{O_4}}} = 0,15\,mol\\
{m_{hh}} = 74,4g \Rightarrow 0,3 \times (2{M_R} + 60) + 0,3 \times (2{M_R} + 96) = 74,4\\
\Rightarrow {M_R} = 23g/mol \Rightarrow R:Natri(Na)\\
b)\\
\% {m_{N{a_2}S{O_4}}} = \dfrac{{0,3 \times 142}}{{74,4}} \times 100\% = 57,26\% \\
\% {m_{N{a_2}C{O_3}}} = 100 - 57,26 = 42,74\%
\end{array}\)