Em tham khảo nha:
\(\begin{array}{l}
a)\\
A{l_4}{C_3} + 12{H_2}O \to 3C{H_4} + 4Al{(OH)_3}\\
b)\\
{m_{A{l_4}{C_3}}} + {m_{{H_2}O}} = {m_{C{H_4}}} + {m_{Al{{(OH)}_3}}}\\
c)\\
{m_{{H_2}O}} = {m_{C{H_4}}} + {m_{Al{{(OH)}_3}}} - {m_{A{l_4}{C_3}}}\\
\Rightarrow {m_{{H_2}O}} = 14,4 + 93,6 - 43,2 = 64,8g\\
{m_{{H_2}O}} \text{ cần dùng }= 64,8 + 64,8 \times 20\% = 77,76g\\
{V_{{H_2}O}} = \dfrac{{77,76}}{1} = 77,76\,ml
\end{array}\)
