Question

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?

Answer 1

The volume of 0.258 M
`Ba(OH)_2` required to react completely with 3.00 g sulfur is 0.364 L

How to calculate the volume of
`Ba(OH)_2`?

First, we shall calculate the volume of
`Ba(OH)_2` that took part in the reaction. This is shown below:

• Mass of S (m) = 3.00 g
• Molar mass of S (M) = 32 g/mol
• Mole of S = m / M = 3 / 32 = 0.094 mole
• Mole of
  `Ba(OH)_2` =?

`H_2(g)\ +\ S(s)\ +\ OH^(-)(aq)\ \rightarrow\ S^(2-)(aq)\ +\ H_2O(l)`

From the balanced equation above,

1 mole of S reacted with 1 mole of
`OH^(-)`

Therefore,

0.094 mole of S will also react with 0.094 mole of
`OH^(-)`

Thus, the mole of
`Ba(OH)_2` that took in the reaction is 0.094 mole

Finally, we shall calculate the volume of
`Ba(OH)_2` that is required for the reaction. Details below:

• Mole of
  `Ba(OH)_2` = 0.094 mole
• Molarity of
  `Ba(OH)_2` = 0.258 M
• Volume of
  `Ba(OH)_2` =?

Volume of
`Ba(OH)_2` = Mole / molarity

= 0.094 / 0.258

= 0.364 L

Answer 2

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 :

`\tt (3~g)/(32~g/mol)=0.09375`

mol OH from eq 1 : (mol S : mol OH = 1 : 2)

`\tt (2)/(1)* 0.09375=0.1875`

mol Ba(OH)₂ from eq 2 : (mol Ba(OH)₂ : mol OH⁻=1 : 2)

`\tt (1)/(2)* 0.1875=0.09375`

volume of Ba(OH)₂ :

`\tt Volume =(mol)/(M)=(0.09375)/(0.258)=0.363 L`