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Question

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The pollutant sulphur dioxide can removed from the air by reaction with calcium carbonate in the presence of oxygen. What mass of calcium carbonate is needed to remove 1 tonne of sulphur dioxide?

2 CaCO3 + 2 SO2 + O2  2 CaSO4 + 2 CO2

15) 5.00 g of hydrated sodium sulphate crystals (Na2SO4.nH2O) gave 2.20 g of anhydrous sodium sulphate on heating to constant mass. Work out the relative molecular mass (Mr) of the hydrated sodium sulphate and the value of n.
Na2SO4.nH2O  Na2SO4 + n H2O

16)5.00 g of a mixture of MgSO4.7H2O and CuSO4.5H2O was heated at 120C until a mixture of the anhydrous salts was formed, which weighed 3.00 g. Calculate the percentage by mass of MgSO4.7H2O in the mixture.

Answer 1

1) the answer is 1.5625 tons
2) n=10 and relative molecular mass is 322
3) the percentage by mass of mgso4.7h2o is 21.5%
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Answer 2

2CaCO3 + 2SO2 + O2 -> 2CaSO4 + 2CO2

Molar Mass of CaCO3 = 100
Molar Mass of SO2 = 64

1 tonne = 1000 kg = 10^6 g

So, moles of SO2 = 10^6/64

Thus mass of CaCO3 = 10^6 x 100/64 g = 1.56 x 10^6 g

15) 5-2.2 = 2.8 g of water was released

This is equal to 2.8/18 = 0.1556 mole of water

Moles of Na2SO4 = 2.2/(2x23+32+64) = 2.2/142 = 0.01549

So, 0.01549 = 0.1556/n

or n = 10 (approx)

16) Let moles of MgSO4 = x

and those of CuSO4 = y

Molar Mass of MgSO4 = 120
Molar Mass of CuSO4 = 159.5

So, 120x + 159.5y = 3

and 120x + 7x * 18 + 159.5y + 5y*18 = 5

Solve for x and y.