Em tham khảo nha :
\(\begin{array}{l}
{\% ^{37}}Cl = 100 - 75 = 25\% \\
{M_{tbCl}} = \dfrac{{35 \times 75 + 37 \times 25}}{{100}} = 35,5dvC\\
{M_{FeC{l_3}}} = 56 + 35,5 \times 3 = 162,5dvC\\
{\% ^{37}}Cl = \dfrac{{25 \times 37 \times 3}}{{162,5}} = 17,08\%
\end{array}\)