sully1993
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Hey, am seriously stuck on these reacting mass calculations, have done quite alot but cant get these around my head :confused: Any help, greatly appreciated

14) The pollutant sulphur dioxide can be 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) 5.00g of hydrated sodium sulphate crystals (Na2SO4.nH2O) gave 2.20g 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 + 2 CO2


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


Thanks alot
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shengoc
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(Original post by sully1993)
Hey, am seriously stuck on these reacting mass calculations, have done quite alot but cant get these around my head :confused: Any help, greatly appreciated

14) The pollutant sulphur dioxide can be 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
This is supposed to be an exercise for you, and you are supposed to go and research for similar scenarios and try to work this out, but anyway, you ought to have done moles and stuff.

First mole of SO2 = mass/RMM, The mass needs to be in grammes, 1 tonne = 10^6 g. Then from the balanced eqn, mole of CaCO3 = mole of SO2(same ratio) = mass of CaCO3/RMM of CaCO3



15) 5.00g of hydrated sodium sulphate crystals (Na2SO4.nH2O) gave 2.20g 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 + 2 CO2[/QUOTE]
Again, work out mole of Na2SO4(anhydrous) using mole = mass/RMM, then work out mole of the hydrated Na2SO4, since the ratio is 1:1, therefore mole of Na2SO4 hydrated = mole of Na2SO4 = 5/RMM of hte hydrated Na2SO4, simply write the RMM of the hydrated one as RMM of Na2SO4 + n(18), work out value of n.


16) 5.00g of a mixture of MgSO4 . 7H2O and CuSO4 . 5H2O was heated at 120 degrees Celsius until a mixture of the anhydrous salts was formed, which weighed 3.00g. Calculate the percentage by mass of MgSO4 . 7H2O in the mixture.[/QUOTE]
This is a bit more complicated, give me a minute or two.


Thanks alot [/QUOTE]
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shengoc
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16) 5.00g of a mixture of MgSO4 . 7H2O and CuSO4 . 5H2O was heated at 120 degrees Celsius until a mixture of the anhydrous salts was formed, which weighed 3.00g. Calculate the percentage by mass of MgSO4 . 7H2O in the mixture.[/QUOTE]
Ok, as promised, i will try to get this question right. Let the initial mass of the hydrated MgSO4 be x, therefore the initial mass of the hydrated Na2SO4 = (5-x) g

Mass of water lost = 5 - 3 = 2g

mole of water lost = 2/18 = 1/9 mole

However, if you construct the decomposition eqn for both the hydrated Mg and Na sulphate, you would get 7 moles of H2O and 5 respectively.

Hence, there are 12 moles of water in total, mole of H20 produced by hydrated MgSO4 = 7/12 x 1/9

Mole of hydrated MgSO4 = 1/7 x mole of water produced
= 1/7 x 7/12 x 1/9 = 1/108 = mass/RMM of the hydrated MgSO4

Therefore mass of hydrated MgSO4 = 1/108 x 246 = 2.278 g

So mass of hydrated NaSO4 = 5 - 2.278 = 2.722 g

Hence the percentage by mass of the Mg... to Na... is like 45.6:54.4.

I think this should be the correct method. Anyone who finds this wrong please quote me, i'd like to know where i'd gone wrong anyway.

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mahek.h1209
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why have u mentioned NaSO4 when its talking about mgso4 and cuso4. i got most of what u did but the last bit, i think is not right
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