Question

Answer 1

\(n_{H2}=\dfrac{V}{22,4}=0,18\left(mol\right)\)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

.0,06...................0,06.........

\(2Al+Ba\left(OH\right)_2+2H_2O\rightarrow Ba\left(AlO_2\right)_2+3H_2\)

.0,12....0,06...............................................0,18...

Mà \(\dfrac{n_{BaO}}{n_{Al}}=\dfrac{3}{7}=\dfrac{0,06}{n_{Al}}\)

\(\Rightarrow n_{Al}=0,14\left(mol\right)\)

\(\Rightarrow n_{Aldu}=n_{Al}-n_{Alpu}=0,14-0,12=0,02\left(mol\right)\)

\(\Rightarrow m=n.M=0,54g\)