PTHH: \(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\uparrow\)
\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)
Ta có: \(n_{Ba}=\dfrac{27,4}{137}=0,2\left(mol\right)=n_{H_2}\)
\(\Rightarrow V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\)
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\(n_{Ba} = \dfrac{27,4}{137} = 0,2(mol)\\ n_{H_2SO_4} = \dfrac{9,8}{98} = 0,1(mol)\)
\(Ba + H_2SO_4 \to BaSO_4 + H_2\)
0,1____0,1____________0,1___(mol)
\(Ba + 2H_2O \to Ba(OH)_2 + H_2\)
0,1____________________0,1____(mol)
Suy ra :
\(n_{H_2} = 0,1 + 0,1 = 0,2(mol)\\ \Rightarrow V_{H_2} = 0,2.22,4 = 4,48(lít)\)
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