Question

24. (15 points total) Skip and Buffy are analytical chemists who work at Sweatshop Analytical Labs. Sweatshop has inexpensive prices for geotechnical and geochemical analytical services, and has a base of high-volume routine work from cement kilns, gravel quarries, and mineral resource companies as clients. One test that Sweatshop offers involves measuring the calcium carbonate content of a specific feed material. The feed material contains a mixture of CaCO3 (s) and other, non-reacting solids such as SiO2 (s). The sample is treated in an excess of HCl(aq); the generated CO2 (g) is collected and dried with a dessicant to remove water vapor, and the pure CO2 (g) is then collected in a balloon of measured V under known P and T conditions. To summarize: Moles of CaCO3 (s) equates to moles CO2 (s). The number of moles of the carbon dioxide determines V, which gives number of moles CO2 under conditions of known P and T. A. (3 points) Write a balanced net ionic equation depicting the chemical process that Skip and Buffy are relying on to report their results. B. (6 points) A 3.00 gram unknown CaCO3 -SiO2 sample produces Vc02 = 0.656 L, when T = 20.0°C and P = 749 Torr (mm Hg). What is the composition of the sample? Please fill in Skip and Buffy's lab report sheet: Mass of CaCO3 (s) found in 3.00 grams mixture = _ Mass of SiO2 (s) found in 3.00 g mixture (by difference) = Percent CaCO3 (8) = –

Answer 1

(A) Net ionic equation :

CaCO₃(s) + 2 HCl(aq) $\rightarrow$ CaCl₂(aq) + H₂O(g) + CO₂(g)

(B)

Number of moles of CO₂ = PV/RT = (749/760) atm× 0.656L/(0.0821atm-L/K.mol × 293.15K) = 0.026862mol

Number of moles of CaCO₃ = number of moles of CO₂

Mass of CaCO₃ = 0.026862mol×100g/mol = 2.6862g

= 2.69g. (Answer)

Mass of SiO₂ = 3.00 - 2.6862 = 0.3138g = 0.314g (Answer)

Percent CaCO₃ = 2.6862/3.00 × 100% = 89.54%

= 89.5 %. (Answer)

(D)

The number of moles of CO₂ calculated would be more and hence more number of moles calculated for CaCO₃. The results would be higher than the correct percent CaCO₃.