Calculate the % of the black oxide ( Fe3O4 ) in the ore . please help :) i cant understand how to do it !
2 F3O4 + 0.5 O2 3 F2O3 Every 2 mole of F3O4 gives 3 moles of F2O3 every 481.1 grams of F3O4 gives 497.1 grams of F2O3 so theoretically every 0.5 grams should give 0.5166 grams, but it gave only 0.411 grams So the percent of Black oxide is sample equals 0.411/0.5166 x 100 79.55% 80%
Supposing Fe3O4 is the only source of iron in the sample: (0.411 g Fe2O3) / (159.6887 g Fe2O3/mol) x (2 mol Fe3O4 / 3 mol Fe2O3) x (231.5333 g Fe3O4/mol) / (0.5 g) 0.8 80% Fe3O4
