I‘m still doing a chemistry page and need help with this question. can someone please explain how to do it.i need a step by step explanation if possible! thank you!
calculate how much fe2o3 can be extracted frm 1000 kg of ore ie 630 kg now from fe2so3 55.8*2 gm(molar mass of fe) of fe could be extracted 159 .6 gm of fe2o3(molar mass of fe2o3)111.6gm of fe 1gm111.6/159.6 630kg(111.6/159.6)*630 440.5 kg
I think you can blame gravity. Wherever the most weight is, that's where you'll land.
I assume it's to avoid landing all at once, much as you're supposed to tuck and roll after a fall.
% by mass of Fe2O3 (mass of pure Fe2O3 / mass of iron ore) x 100% mass of Fe2O3 (pure) (mass of iron produced / MW of iron) x ( stoichiometric mole ratio of Fe2O3 over Fe based on the given chemical reaction x MW of Fe2O3 mass of iron produced 453 g MW of Fe 55.85 stoichiometric mole ratio of Fe2O3/Fe 1/2 MW of Fe2O3 159.7 mass of pure Fe2O3 (453/55.85) x (1/2) x 159.7 mass of pure Fe2O3 647.66 grams %by mass of Fe2O3 (647.66 / 752) x 100 86.13 %
calculate how much fe2o3 can be extracted frm 1000 kg of ore ie 630 kg now from fe2so3 55.8*2 gm(molar mass of fe) of fe could be extracted 159 .6 gm of fe2o3(molar mass of fe2o3)111.6gm of fe 1gm111.6/159.6 630kg(111.6/159.6)*630 440.5 kg
Do your own homework - But. Eliminath the 38% contamination Then workout the ratios of atomic weights for the Fe2O3 Apply the ratio to the 620 kg of ore Presto. assuming a 100% recovery rate - you may need to adjust.
I think you can blame gravity. Wherever the most weight is, that's where you'll land.
I assume it's to avoid landing all at once, much as you're supposed to tuck and roll after a fall.
% by mass of Fe2O3 (mass of pure Fe2O3 / mass of iron ore) x 100% mass of Fe2O3 (pure) (mass of iron produced / MW of iron) x ( stoichiometric mole ratio of Fe2O3 over Fe based on the given chemical reaction x MW of Fe2O3 mass of iron produced 453 g MW of Fe 55.85 stoichiometric mole ratio of Fe2O3/Fe 1/2 MW of Fe2O3 159.7 mass of pure Fe2O3 (453/55.85) x (1/2) x 159.7 mass of pure Fe2O3 647.66 grams %by mass of Fe2O3 (647.66 / 752) x 100 86.13 %
Do your own homework - But. Eliminath the 38% contamination Then workout the ratios of atomic weights for the Fe2O3 Apply the ratio to the 620 kg of ore Presto. assuming a 100% recovery rate - you may need to adjust.
