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 which occurred in the titration is:
HCl + NaOH -> NaCl + H2O
b. To find the number of moles of NaOH required in the titration, we can use the following formula:
moles NaOH = concentration NaOH * volume NaOH solution
moles NaOH = 0.1010 M * 21.33 ml * (1 L / 1000 ml)
moles NaOH = 0.00215963 moles
c. To find the number of moles of HCl in the 10.00 ml solution used in the titration, we can use the same formula as in part b:
moles HCl = concentration HCl * volume HCl solution
moles HCl = 1.421 M * 10.00 ml * (1 L / 1000 ml)
moles HCl = 0.01421 moles
d. The total number of moles in the 250.0 ml solution in the volumetric flask can be calculated using the following formula:
moles in solution = concentration solution * volume solution
moles in solution = (0.01421 moles / 10.00 ml) * 250.0 ml
moles in solution = 0.1776 moles
e. The total number of moles of HCl which were added to the sample of calcium carbonate is the same as the number of moles in the 250.0 ml solution, so it is 0.1776 moles.
f. To determine the number of moles of HCl which reacted with calcium carbonate, we need to use the stoichiometry of the reaction between HCl and calcium carbonate. From the balanced equation in part a, we can see that 1 mole of HCl reacts with 1 mole of calcium carbonate. Therefore, the number of moles of HCl that reacted is also 0.1776 moles.
g. The reaction equation for the reaction between HCl and calcium carbonate is:
HCl + CaCO3 -> CaCl2 + H2O + CO2
h. To find the number of moles of calcium carbonate present in the sample, we can use the molar mass of calcium carbonate and the mass of the impure sample:
moles CaCO3 = mass CaCO3 / molar mass CaCO3
moles CaCO3 = 1.635g / 100.0869 g/mol
moles CaCO3 = 0.01634 moles
i. The mass of calcium carbonate present in the sample can be calculated using the following formula:
mass CaCO3 = moles CaCO3 * molar mass CaCO3
mass CaCO3 = 0.01634 moles * 100.0869 g/mol
mass CaCO3 = 1.635 g
j. The mass % of calcium carbonate in the sample can be calculated using the following formula:
mass % = (mass CaCO3 / mass of impure sample) * 100%
mass % = (1.635 g / 1.635 g) * 100%
mass % = 100%
Therefore, the % purity of calcium carbonate is 100%.
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