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Hydrogen gas is bubbled into a solution of barium hydroxide that has sulfur in it. The unbalanced equation for the reaction that takes place is: H2+S+OH- —-> S2- + H2O
What volume of 0.258 M Ba(OH)2 is required to react completely with 3.00g sulfur?

Hydrogen gas is bubbled into a solution of barium hydroxide that has sulfur in it The unbalanced equation for the reaction that takes place is H2SOH gt S2 H2O W class=

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ardni313

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]

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