re: HSC Chemistry Marathon Archive
a) CaCO3 (aq) + 2HCl(aq) --> CaCl2 (aq) + H2O (l) + CO2 (g)
b)n(NaOH) = cv = 1x0.0112 = 0.0112 mol
c) n(HCl) = cv = 10x0.010=0.1mol,
n(CaCO3) = 1.306/ 40.08+12.01+16x3 mol =0.013 (2dp) [here do I need to keep exact value or use rounded value??]
therefore 0.1-2x0.013=0.074 mols of acid remain
d) 0.026 mols
e) m(Ca) = 1.306x40.08/ 40.08+12.01+16x3 = 0.523g (3dp)
f) (0.523/1.306) x100=39.95%
[woops read q wrong and attempted to find the pH of the final solution at part a.... so what was the use of the dilution and NaOH in this question? I did this wrong D:]A student wished to determine the percentage of calcium carbonate in a shell found at the beach. The clean dry shell, which weighed 1.306g, was placed in a small beaker and 10mL of 5mol/L of hydrochloric acid was added. When the shell had completely dissolved, the resulting solution was transferred to a volumetric flask and the volume made up to 25mL with distilled water. A 10mL sample from this solution required 11.2mL of 1mol/L sodium hydroxide for complete neutralisation.
a) Write a balanced equation for the reaction of calcium carbonate with hydrochloric acid. [1]
b) Calculate the number of moles of NaOH present in the 11.2mL of 1mol/L NaOH solution. [1]
c) How many moles of acid remained in the beaker after the reaction with the shell (before the dilution was made)? [2]
d) How many moles of acid reacted with the shell? [1]
e) What mass of calcium carbonate was present in the shell? [2]
f) What was the percentage of calcium carbonate in the shell? [1]
a) CaCO3 (aq) + 2HCl(aq) --> CaCl2 (aq) + H2O (l) + CO2 (g)
b)n(NaOH) = cv = 1x0.0112 = 0.0112 mol
c) n(HCl) = cv = 10x0.010=0.1mol,
n(CaCO3) = 1.306/ 40.08+12.01+16x3 mol =0.013 (2dp) [here do I need to keep exact value or use rounded value??]
therefore 0.1-2x0.013=0.074 mols of acid remain
d) 0.026 mols
e) m(Ca) = 1.306x40.08/ 40.08+12.01+16x3 = 0.523g (3dp)
f) (0.523/1.306) x100=39.95%