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QUESTION

# Calculate the number of oxygen atoms in 120.23 g of iron (II) phosphate, Fe3(PO4)2?

There are eight moles of oxygen atoms in each mole of "Fe"_3("PO"_4)_2". To answer this question, you must convert the given mass of "Fe"_3("PO"_4)_2" to moles of "Fe"_3("PO"_4)_2". Then multiply the moles of "Fe"_3("PO"_4)_2" times eight mol O/1 mol "Fe"_3("PO"_4)_2" to determine the number of moles of oxygen. Then multiply the moles of oxygen times "6.022 x 10"^23 atoms O.

Given/Known: mass of "Fe"_3("PO"_4)_2" = "120.23g" molar mass of "Fe"_3("PO"_4)_2" = "357.478g/mol" 1 mole "Fe"_3("PO"_4)_2" contains 8 moles oxygen atoms 1 mole O atoms = "6.022 x 10"^23 "atoms"

Unknown: number of oxygen atoms in "120.23g Fe"_3("PO"_4)_2

Solution:

1. Convert given mass of "Fe"_3("PO"_4)_2" to moles.

"120.23g Fe"_3("PO"_4)_2" x "1 mol"/"357.478g" = "0.336328mol Fe"_3("PO"_4)_2"

2. Convert moles "Fe"_3("PO"_4)_2" to moles O.

"0.336328mol" "Fe"_3("PO"_4)_2" x "8 mol O""/1mol Fe"_3("PO"_4)_2"" = "2.690624 mol O"

3. Convert moles O to atoms O.

"2.690624 mol O" x "6.022 x 10"^23 "atoms O""/1 mol O" = "1.6203 x 10"^24""atoms O" (answer rounded to five due to 120.23)

"120.23g" "Fe"_3("PO"_4)_2 contains "1.6203 x 10"^24" "atoms O".