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How many moles of Barium Phosphate Ba2(PO4)2 are needed to have 18.02 gram of oxygen atom ?

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To find the number of moles of Barium Phosphate that contain 18.02 grams of oxygen atoms, we need to determine the molar mass of Barium Phosphate and the number of oxygen atoms in one mole of Barium Phosphate.

The molar mass of Barium Phosphate is:
Ba2(PO4)2 = 2137.33 + 31.974 = 601.89 g/mol

There are 2 oxygen atoms in each PO4 group, and there are 2 PO4 groups in one molecule of Barium Phosphate, so there are a total of 4 oxygen atoms in one molecule of Barium Phosphate.

Since there are 4 oxygen atoms in one molecule of Barium Phosphate, and the molar mass is 601.89 g/mol, there are 18.02/601.89 = 0.0300 mol of Barium Phosphate in 18.02 grams of oxygen atoms.

Therefore, there are 0.0300 moles of Barium Phosphate needed to have 18.02 grams of oxygen atoms.
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