Enough concentrated sulfuric acid solution was added to provide 3.00 grams of sulfuric acid which dissolved 4.00 grams of pure fluorapatite. How much excess sulfuric acid was used? Or, in other words how much sulfuric acid was left after it reacted.
Use your post above as a guide.
1. Write and balance the equation.
2. Convert grams fluorapatite to mols. mols = g/molar mass = ?
3. Use the coefficients in the balanced equation (as I did in the problem above) to convert mols fluorapatite to mols H2SO4 used.
4. Convert mols H2SO4 to grams; i.e., g = mols H2SO4 x molar mass H2SO4 = ? = g H2SO4 used.
5. Subtract g H2SO4 initially - g H2SO4 used = g H2SO4 left over.
Post your work if you get stuck.
To determine the amount of excess sulfuric acid used in the reaction, we need to calculate the amount of sulfuric acid that reacted with the fluorapatite and subtract it from the initial amount of sulfuric acid added.
First, let's determine the number of moles of sulfuric acid that reacted with the fluorapatite:
1. Calculate the molar mass of sulfuric acid (H2SO4):
- The atomic masses of hydrogen (H), sulfur (S), and oxygen (O) are approximately 1 g/mol, 32 g/mol, and 16 g/mol, respectively.
- The molar mass of sulfuric acid is:
(2 × atomic mass of hydrogen) + atomic mass of sulfur + (4 × atomic mass of oxygen) = (2 × 1 g/mol) + 32 g/mol + (4 × 16 g/mol) = 98 g/mol.
2. Convert the mass of sulfuric acid that dissolved in the reaction to moles:
- 4.00 grams of sulfuric acid ÷ 98 g/mol = 0.0408 moles.
Now, let's calculate the amount of excess sulfuric acid used:
1. Determine the molar ratio between sulfuric acid and fluorapatite:
- By observing the balanced equation for the reaction between sulfuric acid (H2SO4) and fluorapatite (Ca5(PO4)3F), we can see that the mole ratio is 5:1. In other words, 5 moles of sulfuric acid reacts with 1 mole of fluorapatite.
2. Calculate the moles of sulfuric acid required to react with the given amount of fluorapatite:
- 0.0408 moles of fluorapatite × (5 moles of sulfuric acid / 1 mole of fluorapatite) = 0.204 moles of sulfuric acid.
3. Calculate the moles of excess sulfuric acid:
- Initial moles of sulfuric acid - moles of sulfuric acid used in reaction = Excess moles of sulfuric acid
- 0.204 moles - 0.0408 moles = 0.1632 moles.
Finally, we can convert the excess moles of sulfuric acid into grams:
1. Multiply the moles of excess sulfuric acid by the molar mass of sulfuric acid:
- 0.1632 moles × 98 g/mol = 16.0056 grams.
Therefore, approximately 16.01 grams of sulfuric acid was left after it reacted with the fluorapatite.