An iron ore was found to contain 92.0% pyrite (fool's gold). Pyrite has the formula FeS2and is 46.5% iron. How many grams of iron are there in 50.8 grams of iron ore?

Question

Chemistry

Aydan

13 August, 14:16

An iron ore was found to contain 92.0% pyrite (fool's gold). Pyrite has the formula FeS2and is 46.5% iron. How many grams of iron are there in 50.8 grams of iron ore?

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Aden Fitzgerald · Answer 1

Firstly calculate the grams in the last 8 percent before moving onto the pyrite section.

50.8x0.08=4.064g

We know that iron ore in this case has 92 percent pyrite which contains 46.5 percent iron so we do 50.8x0.92=46.736g from this we need to find 46.5 percent of the iron content in the 92 percent pyrite section then add this answer to the 8 percent of iron ore we found at the start. 46.736x0.465=21.73224g

21.73224g+4.064=25.79624g of iron ore 25.8g (3sf)