What percentage of the Ba2+ in solution is precipitated as BaCO3(s) if equal volumes of 2.4*10^-3 M Na2CO3(s) and 1.1*10^-3M BaCl2(aq) are mixed
To determine the percentage of Ba2+ that is precipitated as BaCO3(s), we need to calculate how much BaCO3 will be formed and compare it to the amount of Ba2+ in the solution.
First, let's set up the balanced chemical equation for the reaction between Na2CO3 and BaCl2:
Na2CO3(aq) + BaCl2(aq) -> BaCO3(s) + 2NaCl(aq)
From the balanced equation, we can see that the mole ratio between Na2CO3 and BaCO3 is 1:1. This means that if 1 mole of Na2CO3 reacts, it will produce 1 mole of BaCO3.
Now, let's calculate the number of moles of BaCO3 formed:
Number of moles of Na2CO3 = concentration (M) × volume (L)
Number of moles of Na2CO3 = 2.4 × 10^-3 M × volume (L) [since the concentration is given]
Since equal volumes of Na2CO3 and BaCl2 are mixed, the volume of Na2CO3 is the same as the volume of BaCl2. So we can substitute the volume of Na2CO3 for the volume of BaCl2.
Number of moles of Na2CO3 = 2.4 × 10^-3 M × 1L (assuming 1L of each solution is mixed)
Number of moles of Na2CO3 = 2.4 × 10^-3 mol
Since the mole ratio between Na2CO3 and BaCO3 is 1:1, the number of moles of BaCO3 formed is also 2.4 × 10^-3 mol.
The number of moles of Ba2+ in the solution is determined by the concentration of BaCl2 and the volume of BaCl2 used.
Number of moles of Ba2+ = concentration (M) × volume (L)
Number of moles of Ba2+ = 1.1 × 10^-3 M × volume (L) [since the concentration is given]
Again, assuming 1L of each solution is mixed, the number of moles of Ba2+ is:
Number of moles of Ba2+ = 1.1 × 10^-3 mol
Now, let's calculate the percentage of Ba2+ that is precipitated as BaCO3:
Percentage of Ba2+ precipitated = (moles of BaCO3 formed / moles of Ba2+ initially) × 100
Percentage of Ba2+ precipitated = (2.4 × 10^-3 mol / 1.1 × 10^-3 mol) × 100
Percentage of Ba2+ precipitated ≈ 218.2%
Therefore, approximately 218.2% of the Ba2+ in the solution would be precipitated as BaCO3(s).