What percentage of the Ba2+ in solution is precipitated as BaCO3(s) if equal volumes of 2.4*10^-3 M Na2CO3(aq) and 1.1*10^-3M BaCl2(aq) are mixed
I made the equation as Na2CO3(aq)+BaCl2(aq)=BaCO3(s)+2NaCl(aq)
Then im assuming you use an ice table to find the concentration of BaCO3. From there i'm not sure how to find the % of Ba2+
is it 100% because the Ba is only seen in the BaCO3 which is precipitated?
To find the percentage of Ba2+ precipitated as BaCO3(s), you can use stoichiometry and the concentrations of the reactants.
First, let's determine the initial moles of BaCl2 and Na2CO3. Since equal volumes of the solutions are mixed, the number of moles will be the same.
Moles of BaCl2 = concentration (mol/L) × volume (L) = (1.1 × 10^-3 mol/L) × V
Moles of Na2CO3 = concentration (mol/L) × volume (L) = (2.4 × 10^-3 mol/L) × V
Since the coefficients in the balanced equation are 1 for BaCl2 and Na2CO3, the number of moles will remain the same throughout the reaction.
Now, let's calculate the final concentrations after the reaction occurs. According to the reaction stoichiometry, one mole of BaCl2 reacts with one mole of Na2CO3 to produce one mole of BaCO3. This means that the moles of BaCO3 formed will be the same as the initial moles of BaCl2.
Moles of BaCO3 = Moles of BaCl2 = (1.1 × 10^-3 mol/L) × V
To find the concentration of BaCO3, divide the moles of BaCO3 by the volume of the solution after mixing:
Concentration of BaCO3 = (1.1 × 10^-3 mol/L) × V / V = 1.1 × 10^-3 mol/L
Finally, to find the percentage of Ba2+ precipitated as BaCO3, divide the moles of BaCO3 by the initial moles of BaCl2 and multiply by 100:
Percentage of Ba2+ precipitated = (1.1 × 10^-3 mol/L) / (1.1 × 10^-3 mol/L) × 100
Percentage of Ba2+ precipitated = 100%
Therefore, 100% of the Ba2+ in solution will be precipitated as BaCO3(s) when equal volumes of 2.4 × 10^-3 M Na2CO3(aq) and 1.1 × 10^-3 M BaCl2(aq) are mixed.