58% Ca3 (PO4) 2 how many phosphorus are in 10 kg bones?

Ca3(po4)2+SiO2+C=CaSiO3+P4

58% Ca3 (PO4) 2 how many phosphorus are in 10 kg bones?

To determine how many phosphorus atoms are in 10 kg of bones, we first need to calculate the moles of Ca3(PO4)2 in 10 kg of bones and then determine the moles of phosphorus present.

1. Calculate the molar mass of Ca3(PO4)2:
Ca (calcium) has a molar mass of 40.08 g/mol.
P (phosphorus) has a molar mass of 30.97 g/mol.
O (oxygen) has a molar mass of 16.00 g/mol.

For Ca3(PO4)2, we have:
3 Ca: 3 * 40.08 g/mol = 120.24 g/mol
2 P: 2 * 30.97 g/mol = 61.94 g/mol
8 O: 8 * 16.00 g/mol = 128.00 g/mol

Total molar mass of Ca3(PO4)2 = 120.24 g/mol + 61.94 g/mol + 128.00 g/mol = 310.18 g/mol

2. Convert the mass of bones (10 kg) to grams:
1 kg = 1000 g, so 10 kg = 10,000 g.

3. Calculate the number of moles of Ca3(PO4)2:
Moles = Mass (g) / Molar mass (g/mol)
Moles = 10,000 g / 310.18 g/mol = 32.26 moles

4. Determine the number of moles of phosphorus:
Since 1 Ca3(PO4)2 molecule contains 2 P atoms, the number of moles of phosphorus is twice the moles of Ca3(PO4)2. Therefore, 32.26 moles * 2 = 64.52 moles of phosphorus.

5. Convert moles of phosphorus to atoms:
Since one mole of any substance contains Avogadro's number of atoms (6.022 x 10^23 atoms/mol), we can multiply the moles of phosphorus by Avogadro's number to get the number of atoms:
64.52 moles * 6.022 x 10^23 atoms/mol = 3.884 x 10^25 atoms of phosphorus.

So, there are approximately 3.884 x 10^25 phosphorus atoms in 10 kg of bones.