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3 How many moles of phosphate ions are in 1.00 lb of Ca3(PO4)? 1 lb 453.59237 g exactly solve everything step by step
Accepted Answer
Here's how to determine the moles of phosphate ions in 1.00 lb of Ca3(PO4):
1. Convert pounds to grams:
1.00 lb
453.59237 g/lb = 453.59237 g
2. Calculate the molar mass of Ca3(PO4)2:
Ca: 40.08 g/mol
3 = 120.24 g/mol
P: 30.97 g/mol
2 = 61.94 g/mol
O: 16.00 g/mol
8 = 128.00 g/mol
Total molar mass = 120.24 + 61.94 + 128.00 = 310.18 g/mol
3. Calculate the moles of Ca3(PO4)2:
moles = mass / molar mass = 453.59237 g / 310.18 g/mol = 1.462 mol Ca3(PO4)2
4. Determine the moles of phosphate ions (PO4^3-)
There are 2 moles of PO4^3- ions for every 1 mole of Ca3(PO4)2.
Moles of PO4^3- = 1.462 mol Ca3(PO4)2
(2 mol PO4^3- / 1 mol Ca3(PO4)2) = 2.924 mol PO4^3-
Therefore, there are 2.924 moles of phosphate ions (PO4^3-) in 1.00 lb of Ca3(PO4)2.
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