\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(\dfrac{1}{15}\) 0,05 \(\dfrac{1}{30}\)
LTL: \(\dfrac{0,1}{4}>\dfrac{0,05}{3}\)
=> Al dư sau pứ.
hh A gồm: \(\left\{{}\begin{matrix}Al_2O_3\\Al_{dư}\end{matrix}\right.\)
\(Al_2O_3+NaOH\rightarrow NaAlO_2+H_2O\)
1/30 1/30
\(2Al+2H_2O+2NaOH\rightarrow2NaAlO_2+3H_2\)
1/15 1/15
\(V_{NaOH}=\dfrac{n_{NaOH}}{CM_{NaOH}}=\dfrac{1:30+1:15}{2}=0,05\left(l\right)\)
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