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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

removal of SO2

with the reaction equation you see that 2 moles of SO2 correspond to 2 moles CaCO3

Mw: CaCO3 =40+12+3*16=100 for SO2 =32+2*16=64
so 64g of SO2 correspond to 100g CaCO3

1 Tonne = 10^6g

and mass CaCO3 = 100=10^6/64 =1562500g = 1,5625 tonne

15)Na2SO4 has a molecular weight of142
Na2SO4
has a molecular weight of 142=5/2.2=322g
so the mass of nH2O =322-142=180
and knowing masS H2O 18 , you find n=10

16)the mass of the first salt (Mg) is 120 +126 =246
and the mass of the fraction of water is 126/246 =0.512
for the second (Cu) 159.5 +90= 249.5
fraction of water is 90/249.5=0.360

the mass of water in the mixture is 5-3 =2g =0.4

so you have if x is the fraction of Mg salt (then (1-x) is fraction of Cu salt)

0.4 = 0.512x+ 0.36(1-x)

0.4 -0.36= 0.152x

x=0.26 for Mg and 0.74 for Cu

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.