Mr(K2FeO4) = 39*2 + 56 + 16*4 = 198
w(Fe,%) = (Mr(Fe)*100%)/Mr(K2FeO4) = (56*100)/198 = 28,28%
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Mr(K2FeO4) = 39*2 + 56 + 16*4 = 198
w(Fe,%) = (Mr(Fe)*100%)/Mr(K2FeO4) = (56*100)/198 = 28,28%