calculations and reaction
Posted: Wed Apr 29, 2020 1:48 am
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
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