Đáp án:
\(\begin{array}{l}
a)\\
{V_{{O_2}}} = 3,36l\\
{V_{kk}} = 16,8l\\
b)\\
{m_{KMn{O_4}}} = 47,4g\\
c)\\
V = 300
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2\,mol\\
{n_{{O_2}}} = 0,2 \times \dfrac{3}{4} = 0,15\,mol\\
{V_{{O_2}}} = 0,15 \times 22,4 = 3,36l\\
{V_{kk}} = 3,36 \times 5 = 16,8l\\
b)\\
2KMn{O_4} \xrightarrow{t^0} {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,3\,mol\\
{m_{KMn{O_4}}} = 0,3 \times 158 = 47,4g\\
c)\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{A{l_2}{O_3}}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{n_{{H_2}S{O_4}}} = 3{n_{A{l_2}{O_3}}} = 0,3\,mol\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,3}}{1} = 0,3l = 300ml
\end{array}\)
