Final answer:
To precipitate all of the barium from 12.5 cm³ of 0.15 mol/dm³ barium nitrate, 25 cm³ of 0.25 mol/dm³ sodium sulfate solution is needed.The correct answer is:b) 25 cm³
Step-by-step explanation:
In order to determine the volume of 0.25 mol/dm³ sodium sulfate solution needed to precipitate all of the barium from 12.5 cm³ of 0.15 mol/dm³ barium nitrate, we can use the stoichiometry of the reaction.
The balanced chemical equation for the reaction between barium nitrate and sodium sulfate is:
Ba(NO₃)₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaNO₃(aq)
From the equation, we can see that 1 mole of barium nitrate reacts with 1 mole of sodium sulfate to produce 1 mole of barium sulfate. Therefore, the number of moles of barium nitrate is equal to the number of moles of sodium sulfate. Using the equation:
moles = concentration × volume
We can calculate the volume of sodium sulfate solution needed as follows:
Volume of sodium sulfate solution (dm³) = (moles of barium nitrate ÷ concentration of sodium sulfate) = (0.15 mol ÷ 0.25 mol/dm³) = 0.6 dm³ = 600 cm³Therefore, the correct answer is:b) 25 cm³