Em tham khảo nha:
\(\begin{array}{l}
a)\\
{n_{Fe}} = \dfrac{{18,066 \times {{10}^{22}}}}{{6,02 \times {{10}^{23}}}} = 0,3011\,mol\\
b)\\
{n_{hh}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
c)\\
{n_{HCl}} = 0,3 \times 0,3 = 0,09\,mol\\
{n_{{H_2}S{O_4}}} = 0,3 \times 9 = 2,7\,mol
\end{array}\)