Em tham khảo nha :
\(\begin{array}{l}
{n_{MgO}} = \dfrac{{16}}{{40}} = 0,4mol\\
{N_{MgO}} = 0,4 \times 6 \times {10^{23}} = 2,4 \times {10^{23}}\\
{N_{CuO}} = 2{N_{MgO}} = 4,8 \times {10^{23}}\\
{n_{CuO}} = \dfrac{{4,8 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,8mol\\
{m_{CuO}} = 0,8 \times 80 = 64g
\end{array}\)