A mixture of 20.0 grams of CaO and BaO is placed in a 12.5 L vessel at 200oC with excess CO2 . The mixture is allowed to react to completion forming CaCO¬3 and BaCO3 .The pressure in the container after reaction is 220 torr. The solid carbonates are neutralized by adding 650.0 ml of 1.0M HCl at 20oC. The final temperature of the solution after reaction is 21.066 oC. A drop of phenolphthalein is added to the solution and it remains colorless. Assume the specific heat capacity of the solution is 4.18j/goC.
Substance ∆Hf (kJ/mol)
Ba+2(aq) -538.4
Cl- (aq) -167.4
Ca+2(aq) -543.0
H2O (l) -286
CO2(g) -393.5
HCl (aq) -167.2
CaCO3 (s) -1207
BaCO3 (s) -1213
Find the initial pressure of the CO2(g) and mass percents of BaO and CaO in the mixture.
The solution is then titrated with NaOH until the solution turns pink. This requires the addition of 275.0 ml of NaOH. Calculate the molarity of the NaOH.
I am not asking anyone to solve this for me. I just do not know where to start, I have an idea of where to go once I get started, but Im not sure what figures i need to calculate first. Please help. Thanks!
2 answers
do your own work cc boy
This is a humongous problem and requires several steps. I would start with PV = nRT and solve for mols CO2 present after equilibrium the reaction is complete. You will need this eventually so mols CO2 initial - mols used = mols CO2 after equilibrium.
Then write the equation for the reaction and
dHrxn = (n*dH products) - (n*dH reactants) if you need this information; I'm not sure you do.
You will need to solve two simultaneous equations. If you let X = grams BaO and Y = grams CaO, then equation 1 is
X + Y = 20
Equation 2 comes from the reaction with HCl.
BaCO3 + 2HCl ==> BaCl2 + H2O + CO2
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
(1/2 mol BaO) + (1/2 mol CaO) = mols HCl used to neutralize. The problem isn't clear that mols HCl is the exact amount needed to neutralize all of the BaCO3 + CaCO3 but I would make that assumption.
Solve those two equation simultaneously for X and Y, then
%BaO = (X/20)*100 = ?
%CaO = )Y/20)*100 = ?
The very last part for M NaOH is
M NaOH = mols NaOH/L NaOH
Then write the equation for the reaction and
dHrxn = (n*dH products) - (n*dH reactants) if you need this information; I'm not sure you do.
You will need to solve two simultaneous equations. If you let X = grams BaO and Y = grams CaO, then equation 1 is
X + Y = 20
Equation 2 comes from the reaction with HCl.
BaCO3 + 2HCl ==> BaCl2 + H2O + CO2
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
(1/2 mol BaO) + (1/2 mol CaO) = mols HCl used to neutralize. The problem isn't clear that mols HCl is the exact amount needed to neutralize all of the BaCO3 + CaCO3 but I would make that assumption.
Solve those two equation simultaneously for X and Y, then
%BaO = (X/20)*100 = ?
%CaO = )Y/20)*100 = ?
The very last part for M NaOH is
M NaOH = mols NaOH/L NaOH
4 answers