Em tham khảo nha :
\(\begin{array}{l}
a)\\
CaO + 2HN{O_3} \to Ca{(N{O_3})_2} + {H_2}O\\
{m_{HN{O_3}}} = \dfrac{{37,8 \times 20}}{{100}} = 7,56g\\
{n_{HN{O_3}}} = \dfrac{{7,56}}{{63}} = 0,12mol\\
{n_{CaO}} = \dfrac{{{n_{HN{O_3}}}}}{2} = 0,06mol\\
{m_{CaO}} = 0,06 \times 56 = 3,36g\\
b)\\
{n_{Ca{{(N{O_3})}_2}}} = \dfrac{{{n_{HN{O_3}}}}}{2} = 0,06mol\\
{m_{Ca{{(N{O_3})}_2}}} = 0,06 \times 164 = 9,84g\\
C{\% _{Ca{{(N{O_3})}_2}}} = \dfrac{{9,84}}{{3,36 + 37,8}} \times 100\% = 23,9\%
\end{array}\)