Volume of Ba(OH)₂ : 0.363 L
Further explanation
Balanced equation :
H₂(g)+S(s)+2OH⁻(aq)⇒ S²⁻(aq)+2H₂O(l) (eq 1)
Ba(OH)₂⇒Ba²⁺+2OH⁻ (eq 2)
mol S :
[tex]\tt \dfrac{3~g}{32~g/mol}=0.09375[/tex]
mol OH from eq 1 : (mol S : mol OH = 1 : 2)
[tex]\tt \dfrac{2}{1}\times 0.09375=0.1875[/tex]
mol Ba(OH)₂ from eq 2 : (mol Ba(OH)₂ : mol OH⁻=1 : 2)
[tex]\tt \dfrac{1}{2}\times 0.1875=0.09375[/tex]
volume of Ba(OH)₂ :
[tex]\tt Volume =\dfrac{mol}{M}=\dfrac{0.09375}{0.258}=0.363 L[/tex]
