How many milliliters of 0.310 M BaCl2 are needed to react completely with 60.0 mL of 0.140 M NaSO4?
BaCl2 + Na2SO4 (note my correction) ==> BaSO4 + 2NaCl
moles Na2SO4 = M x L = ??
Using the coefficients in the balanced equation, convert moles Na2SO4 to moles BaCl2. (Since the ratio is 1:1 this shouldn't be a problem.)
Now, use M BaCl2= moles BaCl2/LBaCl2 and solve for L BaCl2, then convert to mL.
For mass BaSO4, g = moles x molar mass.
To find out how many milliliters of 0.310 M BaCl2 are needed to react completely with 60.0 mL of 0.140 M NaSO4, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between these two compounds.
The balanced chemical equation for the reaction between BaCl2 and Na2SO4 is:
BaCl2 + Na2SO4 -> BaSO4 + 2NaCl
From the balanced equation, we can see that 1 mole of BaCl2 reacts with 1 mole of Na2SO4.
First, let's calculate the number of moles of Na2SO4 in 60.0 mL:
Molarity of Na2SO4 = 0.140 M
Volume of Na2SO4 = 60.0 mL = 0.0600 L
Number of moles of Na2SO4 = Molarity of Na2SO4 x Volume of Na2SO4
= 0.140 M x 0.0600 L
= 0.00840 moles Na2SO4
Since the balanced equation states that 1 mole of BaCl2 reacts with 1 mole of Na2SO4, we can conclude that 0.00840 moles of BaCl2 is needed to react completely with the given amount of Na2SO4.
Now, we can use the molarity and the number of moles of BaCl2 to calculate the volume needed:
Molarity of BaCl2 = 0.310 M
Number of moles of BaCl2 = 0.00840 moles BaCl2
Volume of BaCl2 = Number of moles of BaCl2 / Molarity of BaCl2
= 0.00840 moles / 0.310 M
= 0.0271 L
Finally, we can convert the volume from liters to milliliters:
Volume of BaCl2 = 0.0271 L x 1000 mL/L
= 27.1 mL
Therefore, 27.1 milliliters of 0.310 M BaCl2 are needed to react completely with 60.0 mL of 0.140 M Na2SO4.