Đáp án:
\(\begin{array}{l}
a)\\
{m_A} = 190g\\
b)\\
\% {m_{NaHC{O_3}}} = 44,21\% \\
\% {m_{N{a_2}C{O_3}}} = 55,79\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
T{N_1}:\\
NaHC{O_3} + NaOH \to N{a_2}C{O_3} + {H_2}O\\
{n_{NaHC{O_3}}} = {n_{NaOH}} = 0,5 \times 1 = 0,5\,mol\\
T{N_2}:\\
2NaHC{O_3} + {H_2}S{O_4} \to N{a_2}S{O_4} + 2C{O_2} + 2{H_2}O\\
N{a_2}C{O_3} + {H_2}S{O_4} \to N{a_2}S{O_4} + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{22,4}}{{22,4}} = 1\,mol\\
\Rightarrow {n_{N{a_2}C{O_3}}} = 1 - 0,5 = 0,5mol\\
{m_A} = 1 \times 84 + 1 \times 106 = 190g\\
b)\\
\% {m_{NaHC{O_3}}} = \dfrac{{1 \times 84}}{{190}} \times 100\% = 44,21\% \\
\% {m_{N{a_2}C{O_3}}} = 100 - 44,21 = 55,79\%
\end{array}\)
