Question

Answer 1

1.

Vì \(V_{H_2\left(TN_1\right)}>V_{H_2\left(TN_2\right)}\) => TN₁ Al dư

- TN₁: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: \(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)

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

Theo PT: \(n_{H_2}=n_{Ba}+3n_{Ba\left(OH\right)_2}=n_{Ba}+3n_{Ba}=4n_{Ba}\)

=> \(n_{Ba}=\dfrac{0,2}{4}=0,05\left(mol\right)\)

- TN₂: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: \(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\) 

            \(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)

Theo PT: \(n_{H_2}=n_{Ba}+\dfrac{3}{2}n_{Al}\)

=> \(n_{Al}=\dfrac{2}{3}.\left(0,5-0,05\right)=0,3\left(mol\right)\)

=> m = 0,05.137 + 0,3.27 = 14,95 (g)