Đáp án:
\(\begin{array}{l}
b)\\
{V_{{O_2}}} = 6,72l\\
c)\\
{m_{{H_2}O}} = 10,8g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
b)\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{10.8}}{{27}} = 0,4mol\\
{n_{{O_2}}} = \dfrac{3}{4}{n_{Al}} = 0,3mol\\
{V_{{O_2}}} = n \times 22,4 = 0,3 \times 22,4 = 6,72l\\
c)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{15,6}}{{22,4}} = 0,696mol\\
2{H_2} + {O_2} \to 2{H_2}O\\
\dfrac{{0,696}}{2} > \dfrac{{0,3}}{2} \Rightarrow {H_2}\text{ dư}\\
{n_{{H_2}O}} = 2{n_{{O_2}}} = 0,6mol\\
{m_{{H_2}O}} = n \times M = 0,6 \times 18 = 10,8g
\end{array}\)