Question

experiment result: The final solution contains 0.0551g of eggshell powder, 30 drops of 2.0M HCl, 12 drops of standardized NaOH solution, and 2 drops of BTB.

Question: Show your calculations for the determination of the percentage of calcium carbonate in eggshells. The calibrated pipet has a drop volume of 0.0183 mL/drop.

Answer 1

2 moles Percentage of calcium carbonate in a eggshell - Reaction involued - Molar mals (CaCO₃) = 100.1 g/mol 2 HCl (aq) + CaCO3(s) - Călt + CO2(g)+ H2O C 1 moles + 2c6-(aq) Hce (aq) + NaOH (aq) + H2O(l) + Na*caq+ ctcaq) - A drop volume of 0.0183 mL/ drop final solution contains.. 0:0551 g eggshell 30 drops .cf 2.0MHCl → 0.5498103 L of 2.0 M. Hel 12 drops of Standardized NaOH = 0.2196X103 L of SOM standardized NaOH solution No. of moles of Naon used to react excess hel = 0.21962103 LX n.o molll = 0.2196x 103 mol = No. of moles of Hce Initial no. of moles of Hel = 0.549x108x 2.0 mol/ = 1098x10-3 mol No. of moles of Hel react with caloa = (1.098 -0.2196) x103 = 0.8784x16² mol No. of moles of CaCO3 reacted = 'moles of Hel sugited - 0.878 4x100.0004392 2 weight of cacoz = 0.0004392 X 100.19 mol = 0.0439 g % of Calog in 0.0551g eggshell = 0.0439 x 100 = 79.80% mol Scanned with CamScanner 0.0551 = 79.90/0