In an experiment, the percentage purity of a sample of calcium carbonate was determined: 50.00ml of 1.421 M HCl was added to 1.635g of impure calcium carbonate. The resulting mixture was made up to 250.0ml in a volumetric flask. 10.00 ml of portions of this solution were titrated against 0.1010M NaOH; on average 21.33 ml of the NaOH solution was required.
a. Write the balanced equation for the reaction which occurred in the titration
b. Find the number of moles of NaOH required in the titration
c. Fid the number of moles of HCl in the 10.00 ml solution used in the titration
d. Determine the total number of moles in the 250.0 ml solution in the volumetric flask (= number of moles left after the reaction with calcium carbonate)
e. Calculate the total number of moles of HCl which were added to the sample of calcium carbonate
f. Determine the number of moles of HCl which reacted with calcium carbonate
g. Give the reaction equation for the reaction between HCl and calcium carbonate
h. How many moles of calcium carbonate were present in the sample
i. Calculate the mass of calcium carbonate present in the sample
j. Determine the mass % of calcium carbonate in the sample (= % purity of calcium oxide)
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a. The balanced equation for the reaction between HCl and calcium carbonate is:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2
b. To find the number of moles of NaOH required in the titration, we can use the relationship between moles and volume:
0.1010 M NaOH = moles NaOH / 0.01000 L
moles NaOH = 0.1010 M * 0.01000 L = 0.00101 moles
c. Since the balanced equation shows that 2 moles of HCl react with 1 mole of NaOH, the number of moles of HCl in the 10.00 ml solution used in the titration can be calculated:
moles HCl = (0.00101 moles NaOH) * (2 moles HCl / 1 mole NaOH) = 0.00202 moles
d. The number of moles in the 250.0 ml solution in the volumetric flask can be calculated by using the dilution equation:
moles = Molarity * Volume
moles HCl = (1.421 M * 50.00 ml) / 1000 ml/L = 0.07105 moles
e. The total number of moles of HCl added to the sample of calcium carbonate is the same as the number of moles in the 250.0 ml solution:
0.07105 moles
f. The number of moles of HCl that reacted with calcium carbonate can be determined from the stoichiometry of the balanced equation. From the equation, we can see that 1 mole of calcium carbonate reacts with 2 moles of HCl:
moles HCl reacted = 0.07105 moles HCl * (1 mole CaCO3 / 2 moles HCl) = 0.03553 moles
g. The reaction equation for the reaction between HCl and calcium carbonate is:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2
h. The stoichiometry of the balanced equation tells us that 1 mole of calcium carbonate reacts with 1 mole of CaCO3:
moles CaCO3 = 0.03553 moles
i. To calculate the mass of calcium carbonate present in the sample, we need to use the molar mass of CaCO3:
mass CaCO3 = moles CaCO3 * molar mass CaCO3
mass CaCO3 = 0.03553 moles * (40.08 g/mol + 12.01 g/mol + (3 * 16.00 g/mol))
mass CaCO3 = 0.03553 moles * 100.09 g/mol = 3.554 g
j. The mass % of calcium carbonate in the sample can be determined by dividing the mass of calcium carbonate by the mass of the impure sample and multiplying by 100:
mass % CaCO3 = (mass CaCO3 / mass impure sample) * 100
mass % CaCO3 = (3.554 g / 1.635 g) * 100 = 217.69% (rounded to 2 decimal places)
Please note that it is not possible to have a mass % greater than 100%. There might be an error in the calculation or the given data.
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