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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?

Respuesta :

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)

There are 25.8 grams of iron present in 50.8 grams of iron ore.

Calculation,

Firstly calculate the grams in the last 8 (100 - 92) percent before moving onto the pyrite section.

50.8 x 0.08 = 4.064 g

Since,iron ore contain 92 percent pyrite

So, 50.8  grams of iron ore contain 50.8 x 0.92 = 46.736 g pyrite

from this we need to find 46.5 percent of the iron content in the 92 percent pyrite section we found at the start.46.736 x 0.465 = 21.73224 g

Then add this answer to the 8 percent of iron ore

21.73224 g +4.064 = 25.79624 g of iron ore 25.8 g

What is pyrite?

  • The pyrite is yellow shiny mineral.
  • This mineral consist iron disulphide
  • It is also known as fool's gold because it resemblance to the gold due to its color.

learn about pyrite,

https://brainly.com/question/13435246

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