An ore contains Fe3O4 and no other iron. The iron in a 46.7 gram sample of the ore is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 25.7 grams. What was the percent Fe3O4 in the sample ore? Answer in %. Please help.
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2 moles of Fe3O4 produce 3 moles of Fe2O3 because there are six iron atoms on both sides of the equation (25.7 g Fe2O3) / (159.6882 g Fe2O3/mol) x (2 mol Fe3O4 / 3 mol Fe2O3) x (231.5326 g Fe3O4/mol) / (45.7 g) 0.54358 54.4% Fe3O4
Weight of Fe3O4 in the sample 2(Fe3O4)/3(Fe2O3) x 25.70 (464/480) x 25.70 24.8433g. Hence %Fe3O4 (24.8433/46.70) x 100 53.20%
231.53 215.53 2Fe 111.69 the mass fraction of Fe in Fe2O3 is 111.69/215.53 so 111.69 / 215.53 * 25.7 13.32 g of iron then Fe% / 100% 13.32 / 46.7 0.285 or 28.5% iron in the ore sample If someone gives you a good answer, please consider giving a best answer. This is the only reward we get.
