sydneymcmahan05 sydneymcmahan05
  • 14-01-2021
  • Chemistry
contestada

How many grams of Fe3+ are present in 2.56 grams of iron(III) iodide?

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bububy
bububy bububy
  • 14-01-2021

Answer:

436.55841 grams.

Explanation:

The molecular formula for Iron(III) Iodide is FeI3. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Iron(III) Iodide, or 436.55841 grams.

You use the 1 / 436.55841 conversion rate. This is different for each element, according to the stoichiometry tables. Iron(II) oxide would be 2/3 of the 1/436.55841 conversion rate, that is, multiplying the fraction to denominator.

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