an iron ore contains 50.0% Fe2O3. How many grams of pure iron could be contained from 200 grams of this ore?
200 g x 0.50 = ?? g Fe2O3.
moles Fe2O3 = ??g/molar mass Fe2O3.
moles Fe = 3 x moles Fe2O3.
g Fe = moles Fe x molar mass Fe.
70 g Fe
Iron ore contain 50% of Fe2O3
200g of iron ore contains 0.5×200= 100g of Fe2O3.
RAM = 160
Fe2 = 112
Grams of pure iron in 100g of Fe2O3 = 112/160×100 = 70g.
To find the number of grams of pure iron that could be contained from 200 grams of the ore, you need to determine the amount of iron in the given ore.
Given that the ore contains 50.0% Fe2O3, this means that 50.0% of the 200 grams is Fe2O3.
To calculate the grams of Fe2O3 in the ore, you can multiply the mass of the ore by the percentage of Fe2O3:
Mass of Fe2O3 = (50.0% / 100) * 200 grams
Mass of Fe2O3 = 0.50 * 200 grams
Mass of Fe2O3 = 100 grams of Fe2O3
Since 1 mole of Fe2O3 contains 2 moles of Fe (iron), you need to convert grams of Fe2O3 to grams of iron. To do this, you need to multiply the mass of Fe2O3 by the ratio of the molar mass of Fe (55.85 g/mol) to the molar mass of Fe2O3 (159.69 g/mol).
Mass of iron = (100 grams of Fe2O3) * (1 mole of Fe / 2 moles of Fe2O3) * (55.85 g / 1 mole)
Mass of iron = (100 grams) * (0.5) * (55.85 g)
Mass of iron = 2792.5 grams
Therefore, from 200 grams of this ore, you could obtain 2792.5 grams of pure iron.