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Chemistry Moles

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?

[1]

(c) How many moles of acid reacted with the shell? (10/1000) * 5 = 0.05ml

[1]

(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!

Reply 1

macpatgh-Sheldon

Original post by username6088380

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 1mol dm ^ - 3 NaOH solution.[1]

(28/1000) * 1 = 0.028mol [correct]

(b) How many moles of acid remained in the beaker after the reaction with the shell?

[1] This is the same as the no of moles of acid that reacted with 0.028 moles of NaOH FROM: NaOH + HCl -----? NaCl + H2O = 0.028

(c) How many moles of acid reacted with the shell? (10/1000) * 5 = 0.05ml [INCORRECT - this was the total added of which 0.028 moles remained as above & reacted with NaOH
ANSWER: 0.05 - 0.028 = 0.022 moles.

[1]

(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 [incorrect cos NOT ALL acid reacted with shell]

ANSWER: {0.022 X Mass No of CaCO3 = 0.022 X 100} /2 [cos 2 moles of acid react with one mole of CaCO3, yeah [see equation in given Q]?
SO = 1.1 g

What was the percentage of calcium carbonate in the shell?

SIMPLE NOW: you can do it - what percentage of 1.216g is 1.1g? Good - well done!

M

PS This Q looks like it is in the wrong place as it is more like GCSE.