Question

200 impure caco3 on heating gives 11.35L co2 gas at stp. find the percentage of calcium in the limestone sample.

Answer 1

Hello Dear.

Here is the answer---

Mass of the impure CaCO₃ = 200 g.
Molar Mass of the CaCO₃ = 100 g/mole.
Molar mass of the copper = 40 g/mole.

Now,
∵ 100 g of CaCO₃ contains 40 g of Calcium.
∴ 1 g of CaCO₃ contains 40/100 g of Calcium.
∴ 200 g of CaCO₃ contains 40 × 2 g of Calcium.
 = 80 g of Calcium.


Now,
% Composition of Calcium in 200 g of CaCO₃ = $\frac{Mass of Ca}{Mass of limestone}$ × 100%
= (80/200) × 100
= 80/2
= 40%.

∴ % of Calcium in the limestone (CaCO₃) sample of 200 grams is 40%.


Hope it helps.

Answer 2

Answer:

A 100 gm sample of impure calcium carbonate is heated, it decomposes to form CO2 gas and calcium oxide. After heating the solid resuide weighted 78g, what was % of calcium carbonate by mass in the original sample?