Expert: George Maxwell - 9/22/2007

Question
I have these questions that I am really stuck on..some help would be greatly appreciated..

1) How much calcium oxide would be formed from 800kg of calcium carbonate?
2) Copper Carbonate reacts with sulphuric acid to give copper sulphate, carbon dioxide and water. What mass of copper sulphate is formed when 225g of copper carbonate is reacted with excess acid?
3) Zinc reacts with copper sulphate to give pure copper. Calculate the mass of copper obtained when 130g of zinc is treated with excess copper sulphate.
4) Barium chloride reacts with sodium suplhate to give a precipiate of insoluble barium sulphate, with sodium chloride as a by-product. Calculate the mass of barium sulphate formed when 2.5cm3 of 0.1M barium chloride is added to excess sodium sulphate.

Thank you so much in advance!

Answer
Hi, and thanks for your questions.

1)
The reaction is:
CaCO3 -> CaO + CO2

The equation is balanced, so one mole of CaCO3 forms one mole of CaO.

One mole of CaCO3 is 100g, and one mole of CaO is 56g. I worked out this using the RAM numbers for each element, which are found on the periodic table: e.g. RAM Ca = 40 and RAM O = 16, so RAM of CaO = 40 + 16 = 56.

So 1kg of CaCO3 will form (using the same ratio) 560g CaO.

Multiplying up, 800 kg of CaCO3 will form (800 x 0.560) = 448 kg CaO. Note that I changed the measurements in to kg during calculation.

2)
Asssuming the equation is balanced (I think it is):
1 mole of Copper Carbonate forms 1 mole of copper sulphate.

1 mole of  copper carbonate weighs 124g, so 225g is 1.814 moles. I worked this out by doing 225 / 124.

If 1 mole CuCO3 makes 1 mole CuS04, then 1.814 moles will make 1.814 moles CuSO4.

Therefore, you just have to work out what 1.814 moles of CuSO4 would weigh:

1 mole is 160g, so 1.814 moles are 1.814 x 160 = 290.24g

3)
Zn + CuSO4 -> ZnSO4 + Cu

1 mole of Zinc weighs 65g. You have 130g which is
(130 / 65)  2 moles. Therefore you make 2 moles of Copper.
1 mole of copper is 64g, so 2 moles is 128g.

I hope this helps, I've left the last question to see if this method works for you, but if you have any other problems, let me know.

Thanks again.