how much of a 1.50M solution sodium sulfate solution in milliliters is required to completely precipitate all the barium in 150.0mL of a 0.250 M barium nitrate solution
Ba(NO3)2 + Na2SO4 ==> BaSO4 + 2NaNO3
millimoles Ba(NO3)2 = mL x M = ?
mmoles Na2SO4 needed = the same since 1 mole Ba(NO3)2 = 1 mole Na2SO4.
mmoles Na2SO4 = mL x M
You know M, solve for mL.
To determine how much of a 1.50M sodium sulfate solution is required to completely precipitate all the barium in a 150.0mL of 0.250M barium nitrate solution, we need to use stoichiometry and the concept of limiting reactants.
First, let's write out the balanced chemical equation for the reaction between sodium sulfate (Na2SO4) and barium nitrate (Ba(NO3)2) to form barium sulfate (BaSO4) and sodium nitrate (NaNO3):
Na2SO4 + Ba(NO3)2 → BaSO4 + 2NaNO3
From the equation, we can see that one molecule of sodium sulfate reacts with one molecule of barium nitrate to produce one molecule of barium sulfate.
Next, we need to calculate the number of moles of barium nitrate in the given solution:
Number of moles = concentration × volume
= 0.250 M × (150.0 mL / 1000 mL)
= 0.0375 moles
According to the balanced equation, we can see that one mole of barium nitrate reacts with one mole of sodium sulfate to produce one mole of barium sulfate. Therefore, we need 0.0375 moles of sodium sulfate to react with the given amount of barium nitrate.
Now, let's calculate the volume of 1.50M sodium sulfate solution required to have 0.0375 moles of sodium sulfate:
Volume (in liters) = moles / concentration
= 0.0375 moles / 1.50 M
= 0.025 L
Since 1 L is equal to 1000 mL, the volume in milliliters is:
Volume (in mL) = 0.025 L × 1000 mL/L
= 25 mL
Therefore, 25 mL of a 1.50M sodium sulfate solution is required to completely precipitate all the barium in 150.0 mL of a 0.250M barium nitrate solution.