Em tham khảo nha:

\(\begin{array}{l}
1)\\
a)\\
{N_{{H_2}S{O_4}}} = n \times 6 \times {10^{23}} = 0,2 \times 6 \times {10^{23}} = 1,2 \times {10^{23}}\\
b)\\
{N_{CuC{l_2}}} = n \times 6 \times {10^{23}} = 0,7 \times 6 \times {10^{23}} = 4,2 \times {10^{23}}\\
2)\\
a)\\
{n_{Fe}} = \dfrac{N}{{6 \times {{10}^{23}}}} = \dfrac{{1,8 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,3\,mol\\
b)\\
{n_{F{e_2}{O_3}}} = \dfrac{N}{{6 \times {{10}^{23}}}} = \dfrac{{2,4 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,4\,mol\\
3)\\
a)\\
{m_{A{l_2}{{(S{O_4})}_3}}} = n \times M = 0,1 \times 342 = 34,2g\\
b)\\
{m_{Fe{{(OH)}_3}}} = n \times M = 0,1 \times 107 = 10,7g\\
c)\\
{n_{S{O_2}}} = \dfrac{N}{{6 \times {{10}^{23}}}} = \dfrac{{0,6 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,1\,mol\\
{m_{S{O_2}}} = n \times M = 0,1 \times 64 = 6,4g\\
4)\\
a)\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{4}{{40}} = 0,1\,mol\\
b)\\
{n_{{H_2}S{O_4}}} = \dfrac{m}{M} = \dfrac{{19,8}}{{98}} = 0,2\,mol\\
c)\\
{n_{{H_2}S}} = \dfrac{N}{{6 \times {{10}^{23}}}} = \dfrac{{4,8 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,8\,mol\\
d)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{11,2}}{{22,4}} = 0,5\,mol\\
5)\\
a)\\
{M_{N{H_4}N{O_3}}} = 14 \times 2 + 1 \times 4 + 16 \times 3 = 80\,dvC\\
b)\\
{M_{KMn{O_4}}} = 39 + 55 + 16 \times 4 = 158\,dvC\\
c)\\
{M_A} = \dfrac{m}{n} = \dfrac{{6,4}}{{0,1}} = 64dvC
\end{array}\)