Tooth enamel consist mainly of hydroxyapatite (Ca5(PO4)3(OH). When tine (II) Flouride is added to toothpaste it reacts with the enamel to product a more decay-resistant material fluorapatite (Ca(PO4)3F. The by-products of this reaction are tin (II) oxide and water. What mass of hydroxypatite can be converted to fluoroapatite by reaction with 0.115 grams of tin (II) fluoride? Write a balanced reaction first.

Note you have the wrong formula for fluorapatite. It is Ca5(PO4)3F.

2Ca5(PO4)3(OH) + SnF2 ==> 2Ca5(PO4)3F + SnO + H2O

This is just like the problem I did for you last night. These stoichiometry problems are done the same way. A four step process.
1. Write and balance the equation. I've done that for you above.

2. Convert what you have (in this case SnF2) to mols. mols = grams/molar mass = ?

3. Convert mols of what you have (mols of SnF2) to mols of what you want (mols of hydroxyapatite). You do this by using the coefficients in the balanced equation.

4. Now convert mols of what you want into grams. grams Ca5(PO4)3(OH) = mols x molar mass

Post your work if you run into trouble.

Thanks

The balanced equation for the reaction between hydroxyapatite and tin (II) fluoride is:

5 Ca5(PO4)3(OH) + 4 SnF2 → Ca5(PO4)3F + 4 SnO + 6 H2O

From the equation, we can see that for every 4 moles of SnF2, 5 moles of Ca5(PO4)3(OH) react to form 1 mole of Ca5(PO4)3F.

Now, we can calculate the number of moles of SnF2 from the given mass:

molar mass of SnF2 = atomic mass of Sn + 2*(atomic mass of F)
= 118.71 g/mol

moles of SnF2 = mass of SnF2 / molar mass of SnF2
= 0.115 g / 118.71 g/mol
≈ 0.00097 mol

According to the balanced equation, for every 4 moles of SnF2, 5 moles of Ca5(PO4)3(OH) will react.

So, the number of moles of Ca5(PO4)3(OH) can be calculated as:

moles of Ca5(PO4)3(OH) = (5/4) * moles of SnF2
= (5/4) * 0.00097 mol
= 0.0012 mol

Finally, we can calculate the mass of hydroxyapatite that can be converted:

mass of hydroxyapatite = moles of Ca5(PO4)3(OH) * molar mass of Ca5(PO4)3(OH)
= 0.0012 mol * (5 * atomic mass of Ca + 3 * atomic mass of P + 12 * atomic mass of O + atomic mass of H)
= 0.0012 mol * (5 * 40.08 + 3 * 30.97 + 12 * 16.00 + 1.01)
≈ 0.99 grams

Therefore, approximately 0.99 grams of hydroxyapatite can be converted to fluorapatite by reacting with 0.115 grams of tin (II) fluoride.

To determine the mass of hydroxyapatite that can be converted to fluorapatite by the reaction with 0.115 grams of tin (II) fluoride (SnF2), we need to first write the balanced reaction equation:

3Ca5(PO4)3(OH) + 10SnF2 → 2Ca(PO4)3F + 10SnO + 6H2O

From the balanced equation, we can see that for every 10 moles of SnF2, two moles of Ca5(PO4)3(OH) are required to form two moles of Ca(PO4)3F.

Now, let's calculate the number of moles of SnF2:

Molar mass of SnF2 = atomic mass of Sn + (2 * atomic mass of F)
Molar mass of SnF2 = (118.71 g/mol) + (2 * 18.998 g/mol)
Molar mass of SnF2 = 156.706 g/mol

Number of moles of SnF2 = mass of SnF2 / molar mass of SnF2
Number of moles of SnF2 = 0.115 g / 156.706 g/mol
Number of moles of SnF2 ≈ 0.000734 mol

According to the balanced equation, for every 10 moles of SnF2, 2 moles of Ca5(PO4)3(OH) are required. Thus, the number of moles of Ca5(PO4)3(OH) is given by the ratio:

Number of moles Ca5(PO4)3(OH) = (0.000734 mol SnF2 / 10) * 2
Number of moles Ca5(PO4)3(OH) = 0.000147 mol

Finally, we can calculate the mass of Ca5(PO4)3(OH):

Molar mass of Ca5(PO4)3(OH) = (5 * atomic mass of Ca) + (3 * atomic mass of P) + atomic mass of O + atomic mass of H
Molar mass of Ca5(PO4)3(OH) = (5 * 40.08 g/mol) + (3 * 30.97 g/mol) + 16.00 g/mol + 1.008 g/mol
Molar mass of Ca5(PO4)3(OH) ≈ 502.31 g/mol

Mass of Ca5(PO4)3(OH) = number of moles Ca5(PO4)3(OH) * molar mass of Ca5(PO4)3(OH)
Mass of Ca5(PO4)3(OH) = 0.000147 mol * 502.31 g/mol
Mass of Ca5(PO4)3(OH) ≈ 0.0738 g

Therefore, approximately 0.0738 grams of hydroxyapatite can be converted to fluoroapatite by reacting with 0.115 grams of tin (II) fluoride.