calculations and reaction
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glebyych
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Joined: 28 Apr 2020
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calculations and reaction


Hello, could you please check if my calculations are right? we had to calculate amount of reagents to make 5g of (k4fecn6). I found that there is a reaction between
Fecl2+6kcn---> k4fecn6+2kcl, Im stuck here, no information on should we first make a solution of fecl2 and same with kcn or rather use them as Krystals?



FeCl2+6KCN-> K4[Fe(CN)6] + 2KCl

m (K4[Fe(CN)6]) = 5 g

KCN:
368 g – 5 g(K4[Fe(CN)6])
6 * 65 g – x1 g (KCN) x1 = m (KCN) = 5,3 g





FeCl2:

368 g – 5 g (K4[Fe(CN)6])
127 g – x2 g (FeCl2) x2 = m (FeCl2) = 1,7 g

127 g – 199 g (FeCl2 * 4H2O)
1,7 g – x3 g x3 = m (FeCl2 * 4H2O) =2,7
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ChenBeier
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It is correct, but don't do this experiment too dangerous. Normally two solutions are made up and mixed together.
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