## calculations and reaction

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glebyych
Newbie
Posts: 2
Joined: Tue Apr 28, 2020 2:15 pm

### 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 â€“ 5 g(K4[Fe(CN)6])
6 * 65 g â€“ x1 g (KCN) x1 = m (KCN) = 5,3 g

FeCl2:

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

127 g â€“ 199 g (FeCl2 * 4H2O)
1,7 g â€“ x3 g x3 = m (FeCl2 * 4H2O) =2,7
ChenBeier
Distinguished Member
Posts: 432
Joined: Wed Sep 27, 2017 7:25 am
Location: Berlin, Germany
It is correct, but don't do this experiment too dangerous. Normally two solutions are made up and mixed together.