Em tham khảo nha :
\(\begin{array}{l}
NaOH + N{H_2}C{H_2}COOH \to N{H_2}C{H_2}COONa + {H_2}O\\
NaOH + {H_2}NCH(C{H_3})COOH \to {H_2}NCH(C{H_3})COONa + {H_2}O\\
hh:N{H_2}C{H_2}COOH(a\,mol),{H_2}NCH(C{H_3})COOH(a\,mol)\\
{n_{NaOH}} = 0,031 \times 1 = 0,031mol\\
\left\{ \begin{array}{l}
75a + 89b = 2,549\\
a + b = 0,031
\end{array} \right.\\
\Rightarrow a = 0,015;b = 0,016\\
{n_{N{H_2}C{H_2}COONa}} = {n_{Glyxin}} = 0,015mol\\
{m_{N{H_2}C{H_2}COONa}} = 0,015 \times 97 = 1,455g\\
{n_{{H_2}NCH(C{H_3})COONa}} = {n_{Alanin}} = 0,016mol\\
{m_{{H_2}NCH(C{H_3})COONa}} = 0,016 \times 123 = 1,968g\\
m = 1,968 + 1,455 = 3,423g
\end{array}\)
