Em tham khảo nha :
\(\begin{array}{l}
a)\\
{n_{Ba{{(OH)}_2}}} = \dfrac{{150}}{{171}} = \dfrac{{50}}{{57}}mol\\
{N_{Ba{{(OH)}_2}}} = \dfrac{{50}}{{57}} \times 6 \times {10^{23}} = 5,263 \times {10^{23}}\text{(Phân tử )}\\
b)\\
{n_{N{a_2}C{O_3}}} = \dfrac{{50}}{{106}} = \dfrac{{25}}{{53}}mol\\
{N_{N{a_2}C{O_3}}} = \dfrac{{25}}{{53}} \times 6 \times {10^{23}} = 2,8302 \times {10^{23}}\text{(Phân tử )}\\
c)\\
{n_{{K_2}O}} = \dfrac{{15}}{{94}} = 0,16mol\\
{N_{{K_2}O}} = 0,16 \times 6 \times {10^{23}} = 9,6 \times {10^{22}}\text{(Phân tử )}\\
d)\\
{n_{{P_2}{O_5}}} = \dfrac{{250}}{{142}} = 1,76mol\\
{N_{{P_2}{O_5}}} = 1,76 \times 6 \times {10^{23}} = 1,056 \times {10^{24}}\text{(Phân tử )}
\end{array}\)
