Giúp mik gấp vssss mik sắp đi hc rồi

Em tham khảo nha:

\(\begin{array}{l}
a)\\
4Al{(N{O_3})_3} \to 2A{l_2}{O_3} + 12N{O_2} + 3{O_2}\\
b)\\
{n_{Al{{(N{O_3})}_3}}} = \dfrac{{1,5 \times {{10}^{21}}}}{{6 \times {{10}^{23}}}} = 0,0025\,mol\\
{n_{{O_2}}} = 0,0025 \times \dfrac{3}{4} = 0,001875\,mol\\
{A_{{O_2}}} = 0,001875 \times 6 \times {10^{23}} = 1,125 \times {10^{21}}\\
c)\\
{n_{A{l_2}{O_3}}} = \dfrac{3}{2} \times 0,001875 = 0,0028125\,mol\\
{m_{A{l_2}{O_3}}} = 0,0028125 \times 102 = 0,286875g
\end{array}\)