If 252mL of 2.5molar HCl solution is added to
343 mL of 2.8 molar Ba(OH)2 solution, what
will be the molarity of BaCl2 in the resulting
solution?
mmols HCl = 252 mL x 2.5M = about 630
mmols Ba(OH)2 = 343 x 2.8 = about 960
but you should confirm all of the calculations Some are estimates.
2HCl + Ba(OH)2 ==> BaCl2 + 2H2O
First, which is the limiting reagent (LR)
Convert mmols HCl to mmols BaCl2. That's 630 x (1 mol BaCl2/2 mol HCl) = about 315 mmols BaCl2.
Convert mmols Ba(OH)2 to mmols BaCl2. That's 960 x (1 mol BaCl2/1 mol Ba(OH)2) = 960 mmols BaCl2.
You see that the two values don't agree (the usual case in LR problems) and one must be wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the LR.
So we have 315 mmols or 0.315 mols BaCl2 formed. The volume of the solution is 252 mL + 343 mL = 595 mL or 0.595 L. Then M = mols/L solution.
To find the molarity of BaCl2 in the resulting solution, we can use the principles of stoichiometry.
First, we need to determine the number of moles of HCl and Ba(OH)2 solutions that are being mixed. We can calculate the moles using the formula:
moles = concentration (M) x volume (L)
For HCl:
moles of HCl = 2.5 M x 0.252 L = 0.63 moles
For Ba(OH)2:
moles of Ba(OH)2 = 2.8 M x 0.343 L = 0.9574 moles
Now, let's look at the balanced equation for the reaction between HCl and Ba(OH)2:
2 HCl + Ba(OH)2 -> BaCl2 + 2 H2O
From the balanced equation, we can see that 2 moles of HCl react with 1 mole of Ba(OH)2 to produce 1 mole of BaCl2.
Using the moles calculated earlier, we can determine the limiting reagent, which is the reagent that will be completely consumed first. The limiting reagent in this case is HCl because we have fewer moles of HCl compared to the stoichiometric ratio.
Since 2 moles of HCl react to produce 1 mole of BaCl2, the number of moles of BaCl2 produced will also be 0.63 moles. However, we need to find the concentration (molarity) of BaCl2 in the resulting solution.
To find the molarity, we'll divide the moles of BaCl2 by the total volume of the resulting solution.
Total volume = volume of HCl solution + volume of Ba(OH)2 solution
Total volume = 0.252 L + 0.343 L = 0.595 L
Now we can calculate the molarity:
molarity of BaCl2 = moles of BaCl2 / total volume
molarity of BaCl2 = 0.63 moles / 0.595 L = 1.06 M
Therefore, the molarity of BaCl2 in the resulting solution is 1.06 M.