Crystallization 10,000 kg of a saturated solution of Na 2 CO 3 at 30ºC is availa
ID: 482900 • Letter: C
Question
Crystallization
10,000 kg of a saturated solution of Na2CO3 at 30ºC is available. You want to crystallize from this solution 3,000 kg of Na2CO310H2O (10 molecules of water are associated with one molecule of Na2CO3) without any accompanying water. Based on the solubility data provided below, calculate the temperature to which the solution should be cooled to achieve the desired crystallization.
Temp (C) Solubility (g Na2CO3 / 100 g H2O)
0 7
10 12.5
20 21.5
30 38.8
Explanation / Answer
m = 10^4 kg
m = 3*10^3
MWofNa2CO3 = 105.9888
MW of Na2CO3*10H2O = 105.9888 + 10*18 = 285.9888 g of Na2CO3*10H2O
%mass of Na2CO3 in hydrate = 105.9888 /285.9888 = 0.3706 * 100% = 37.06%
so...
of 3000 kg, 0.3706*3000 = 1111.8 kg of Na2CO3 must be obtained from the 10 tonnes
Change in mass = (38.3-21.5) = 16.8 g of salt per 100 g of H2O
the composition of 30°C mix:
38.8 g of Na2CO3 and 100 g of H2O, so
x-Na2CO3 = 38.8/(38.8+100) = 0.279538
mass of water = 10000*(1-0.279538) = 7204.62 kg of water
a)
30°C to 20°C:
38.8-21.5 = 17.3 g per 100 g of water... = 17.3 g of salt will crystalise per 0.1 kg of water
total water = 7204.62
so:
17.3/0.1*7204.62 = 1246,399.26 g of salt will be procduced
1.246 kg of salt we need 1111.8 kg of sat..
not enough, now try 30° to 10C
30 t 10 =38.8-12.5 = 26.3
26.3/0.1*7204.62 = 1894815.06 g of salt will be procduced
1894.81506 kg of salt we need 1111.8 kg of sat..
so this will do, cool do 10°C
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