Question

Answer 1

a) \(n_{O_2}=\dfrac{8}{32}=0,25\left(mol\right)\)

=> \(n_{H_2}=0,25\left(mol\right)\)

PTHH: Ba + 2H₂O --> Ba(OH)₂ + H₂

          0,25<-------------0,25<---0,25

=> a = m_Ba = 0,25.137 = 34,25 (g)

=> b = m_Ba(OH)2 = 0,25.171 = 42,75 (g)

b) m_dd = 100 + 34,25 - 0,25.2 = 133,75 (g)

Answer 2

VH2 = VO2 

=> nH2 = nO2 = 8/32 = 0,25 (mol)

PTHH: Ba + 2H2O -> Ba(OH)2 + H2

Mol: 0,25 <--- 0,5 <--- 0,25 <--- 0,25

mBa = 0,25 . 137 = 34,25 (g)

mBa(OH)2 = 0,25 . 171 = 42,75 (g)

mddBa(OH)2 = 42,75 + 100 - 0,25 . 2 = 142,25 (g)

 

Answer 3

a.\(n_{O_2}=\dfrac{8}{32}=0,25mol\)

Khí thoát ra là H2

\(\Rightarrow n_{H_2}=0,25mol\)

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

0,25                    0,25        0,25  ( mol )

\(m_{Ba}=0,25.137=34,25g\)

\(m_{Ba\left(OH\right)_2}=0,25.171=42,75g\)

b.\(m_{Ba\left(OH\right)_2}=42,75+100-\left(0,25.2\right)=142,25g\)

 

Answer 4

nH2 = nO2 = 8 : 22,4 = 5/14 (mol) 
Ba + 2H2O ---> Ba(OH)2 +  H2
5/14<------------<5/14<--------5/14 (mol) 
=> a = mBa = 5/14 . 137 = 48,9 ( g) 
=> b = mBa(OH)2 = 5/14 . 171 = 61,07 (g) 
nH2O = 100 : 2 = 50 (g) 
pthh : Ba + 2H2O ---> Ba(OH)2 +H2 
                    50--------> 25 (mol) 
-> mBa(OH)2 = 25 . 171 = 4275 (g)