How many milliliters of 0.170M BaCl 2 are needed to react completely with 45.0mL of 0.260M Na 2 SO 4 ?
jhbk
To find the answer, we need to use the balanced chemical equation and stoichiometry.
The balanced chemical equation for the reaction between BaCl2 and Na2SO4 is:
BaCl2 + Na2SO4 -> BaSO4 + 2NaCl
From the equation, we can see that one mole of BaCl2 reacts with one mole of Na2SO4.
First, let's find the number of moles of Na2SO4 in 45.0 mL of 0.260M Na2SO4:
Molarity (M) = moles/volume (L)
moles = Molarity * volume
moles of Na2SO4 = 0.260 M * 0.045 L = 0.0117 moles
Since the stoichiometry of BaCl2 to Na2SO4 is 1:1, we can determine that 0.0117 moles of BaCl2 are required to react completely with 0.0117 moles of Na2SO4.
Now, let's calculate the volume of 0.170M BaCl2 needed:
volume (L) = moles/Molarity
volume of BaCl2 = 0.0117 moles / 0.170 M = 0.069 L or 69.0 mL
Therefore, you would need 69.0 milliliters of 0.170M BaCl2 to react completely with 45.0 milliliters of 0.260M Na2SO4.