Em tham khảo nha :

\(\begin{array}{l}
1)\\
{N_{S{O_2}}} = 0,6 \times 6 \times {10^{23}} = 3,6 \times {10^{23}}\text{(Phân tử)}\\
{N_{FeO}} = 2 \times 6 \times {10^{23}} = 1,2 \times {10^{24}}\text{(Phân tử)}\\
{N_{C{l_2}}} = 1,5 \times 6 \times {10^{23}} = 9 \times {10^{23}}\text{(Phân tử)}\\
{N_{N{a_2}S{O_4}}} = 0,5 \times 6 \times {10^{23}} = 3 \times {10^{23}}\text{(Phân tử)}\\
{N_{Ba{{(N{O_3})}_2}}} = 1,5 \times 6 \times {10^{23}} = 9 \times {10^{23}}\text{(Phân tử)}\\
2)\\
{m_{N{a_2}O}} = 0,2 \times 62 = 12,4g\\
{m_{NaCl}} = 0,5 \times 58,5 = 29,25g\\
{m_{{C_6}{H_{12}}{O_6}}} = 0,05 \times 180 = 9g\\
3)\\
{V_{CO}} = 0,15 \times 22,4 = 3,36l\\
{V_{{N_2}}} = 0,3 \times 22,4 = 6,72l\\
{V_{{N_2}{O_5}}} = 1,2 \times 22,4 = 26,88l\\
4)\\
{M_{MgO}} = 24 + 16 = 40dvC\\
{M_{HCl}} = 1 + 35,5 = 36,5dvC\\
{M_{NaCl}} = 23 + 35,5 = 58,5dvC\\
{M_{KOH}} = 39 + 1 + 16 = 56dvC\\
5)\\
{n_{HCl}} = \dfrac{{1,44 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,24mol\\
{n_{N{a_2}S}} = \dfrac{{24 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 4mol\\
6)\\
{n_{C{O_2}}} = \dfrac{{1,5 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,25mol\\
{n_{{O_2}}} = {n_{C{O_2}}} = 0,25mol\\
{N_{{O_2}}} = 0,25 \times 6 \times {10^{23}} = 1,5 \times {10^{23}}\text{(Phân tử)}
\end{array}\)