A student wished to determine the percentage of calcium carbonate present in a small shell found on a local beach. The cleaned, dry shell, which weighed 1.216 g, was placed in a small beaker and 10cm^3 of 5.00 dm ^ - 3 hydrochloric acid added. After the shell had completely dissolved the resulting solution was carefully transferred to a volumetric flask and the volume made up to 25cm ^ 3 with distilled water. This solution required 28cm ^ 3 of 1.00 mol dm ^ - 3 sodium hydroxide solution for complete neutralisation.
The equation for the reaction between hydrochloric acid and calcium carbonate is: CaC O 3(s) + 2HCl (3q) CaCl 2(aq) +CO 2(n) +H 2 O (0)
(a) Calculate the number of moles of NaOH present in 28cm ^ 3 of 1moldm ^ - 3 NaOH solution.[1]
(28/1000) * 1 = 0.028mol
(b) How many moles of acid remained in the beaker after the reaction with the shell?
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(c) How many moles of acid reacted with the shell? (10/1000) * 5 = 0.05ml
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(d) What mass of calcium carbonate was present in the shell?
[1]
0.05/2 = 0.025 mass =0.025*(40.1+12+(16*3))=2.5g
What was the percentage of calcium carbonate in the shell?
[1]
Hi can anyone spot where i went wrong for d as the mass is more than the mass of the shell 🤦*♀️ and also answer the other parts. Thank you!