Em tham khảo nha:

\(\begin{array}{l}
1)\\
a)\,{n_S} = \dfrac{m}{M} = \dfrac{{16}}{{32}} = 0,5\,mol\\
b)\,\,{n_{Fe}} = \dfrac{m}{M} = \dfrac{{28}}{{56}} = 0,5\,mol\\
c)\,{n_{CaO}} = \dfrac{m}{M} = \dfrac{{14}}{{56}} = 0,25\,mol\\
d)\,{n_{C{O_2}}} = \dfrac{m}{M} = \dfrac{{66}}{{44}} = 1,5\,mol\\
e)\,{n_C} = \dfrac{A}{N} = \dfrac{{3 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,5\,mol\\
f){n_{{H_2}O}} = \dfrac{A}{N} = \dfrac{{9 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 1,5\,mol\\
2)\\
a)\,{A_S} = n \times N = 0,25 \times 6 \times {10^{23}} = 1,5 \times {10^{23}} \text{ nguyên tử } \\
b)\,{A_{Al}} = n \times N = 0,65 \times 6 \times {10^{23}} = 3,9 \times {10^{23}} \text{ nguyên tử }\\
c)\,\\
{n_{CuO}} = \dfrac{m}{M} = \dfrac{{40}}{{80}} = 0,5\,mol\\
{A_{CuO}} = n \times N = 0,5 \times 6 \times {10^{23}} = 3 \times {10^{23}} \text{ phân tử }\\
d)\\
{A_N} = n \times N = 0,375 \times 6 \times {10^{23}} = 2,25 \times {10^{23}} \text{ nguyên tử }\\
e)\\
{n_{{H_2}S{O_4}}} = \dfrac{m}{M} = \dfrac{{49}}{{98}} = 0,5\,mol\\
{A_{{H_2}S{O_4}}} = n \times N = 0,5 \times 6 \times {10^{23}} = 3 \times {10^{23}} \text{ phân tử }\\
e)\\
{A_{{O_2}}} = n \times N = 1,5 \times 6 \times {10^{23}} = 9 \times {10^{23}} \text{ phân tử }
\end{array}\)