902.34 grams of Ba3(PO3)2   =  ____________________________  molecules Ba3(PO3)2 Remember that there are 6.02 x 1023 molecules in a mole.

First, we need to find the molar mass of Ba3(PO3)2.

The molar mass of Ba (Barium) is about 137.33 g/mol, P (Phosphorus) is about 30.97 g/mol, and O (Oxygen) is about 16.00 g/mol.

So, the molar mass of Ba3(PO3)2 is (137.33 g/mol * 3) + 2 * [(30.97 g/mol * 1) + (16.00 g/mol * 3) * 2] = 811.99 g/mol.

Next, we need to convert grams to moles. We do this by dividing the mass of Ba3(PO3)2 by its molar mass.

So, 902.34 g ÷ 811.99 g/mol = 1.11 moles of Ba3(PO3)2.

Finally, we convert moles to molecules using Avogadro's number (6.02 x 10^23 molecules/mole).

So, 1.11 moles * 6.02 x 10^23 molecules/mole = 6.68 x 10^23 molecules of Ba3(PO3)2.