Answer : BaS will be the precipitate which will be formed.
Explanation : When all the three solutions namely; are mixed together a white precipitate of BaS
is formed as a product in the solution along with the soluble by product of Ammonium nitrate which is
<u>Answer:</u> The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(1)
<u>For NaOH:</u>
Initial molarity of NaOH solution = 0.195 M
Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
<u>For sulfuric acid:</u>
Initial molarity of sulfuric acid solution = 0.248 M
Volume of solution = 20.0 mL = 0.020 L
Putting values in equation 1, we get:
The chemical equation for the reaction of NaOH and sulfuric acid follows:
By Stoichiometry of the reaction:
2 moles of NaOH reacts with 1 mole of sulfuric acid
So, moles of KOH will react with =
of sulfuric acid
As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.
Thus, NaOH is considered as a limiting reagent because it limits the formation of product.
Excess moles of sulfuric acid =
By Stoichiometry of the reaction:
2 moles of KOH produces 1 mole of sodium sulfate
So, moles of KOH will produce =
of sodium sulfate
Moles of sodium sulfate =
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
Moles of excess sulfuric acid =
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
Moles of NaOH remained = 0 moles
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.