Em tham khảo nha:

\(\begin{array}{l}
15)\\
{n_{CaC{O_3}}} = \dfrac{{25}}{{100}} = 0,25\,mol\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,25\,mol\\
{m_{{H_2}O}} = 13,25 - 0,25 \times 44 = 2,25g\\
{n_{{H_2}O}} = \dfrac{{2,25}}{{18}} = 0,125\,mol\\
{n_{{H_2}O}} < {n_{C{O_2}}} \Rightarrow A:{C_n}{H_{2n - 2}}\\
{n_A} = 0,25 - 0,125 = 0,125\,mol\\
n = \dfrac{{{n_{C{O_2}}}}}{{{n_A}}} = \dfrac{{0,25}}{{0,125}} = 2 \Rightarrow A:{C_2}{H_2}\\
16)\\
\text{ Gọi a là số mol của $CO_2$ }\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = a\,mol\\
{m_{CaC{O_3}}} - {m_{C{O_2}}} = 25,2g \Leftrightarrow 100a - 44a = 25,2\\
 \Rightarrow a = 0,45\,mol\\
m = m = 5,4g\\
{n_{{H_2}O}} = \dfrac{{5,4}}{{18}} = 0,3\,mol\\
{n_{{H_2}O}} < {n_{C{O_2}}} \Rightarrow A:{C_n}{H_{2n - 2}}\\
{n_A} = 0,45 - 0,3 = 0,15\,mol\\
n = \dfrac{{{n_{C{O_2}}}}}{{{n_A}}} = \dfrac{{0,45}}{{0,15}} = 3 \Rightarrow A:{C_3}{H_4}
\end{array}\)