contestada

Hydroxyapatite, Ca5(PO4)3(OH), is the main mineral component of dental enamel, dentin, and bone, and thus has many medical uses. Coating it on metallic implants (such as titanium alloys and stainless steels) helps the body accept the implant. In the form of powder and beads, it is used to fill bone voids, which encourages natural bone to grow into the void. Hydroxyapatite is prepared by adding aqueous phosphoric acid to a dilute slurry of calcium hydroxide. (a) Write a balanced equation for this preparation. (b) What mass of hydroxyapatite could form from 100. g of 85% phosphoric acid and 100. g of calcium hydroxide?

Respuesta :

superman1987

Answer:

(a) 5Ca(OH)₂ + 3H₃PO₄ = Ca₅(PO₄)₃(OH) + 9H₂O.

(b) 135.62 g.

Explanation:

(a) Write a balanced equation for this preparation.

  • The balanced equation for

5Ca(OH)₂ + 3H₃PO₄ = Ca₅(PO₄)₃(OH) + 9H₂O,

that 5 mol of Ca(OH)₂ react with 3 mol of H₃PO₄ to produce 1 mol of Ca₅(PO₄)₃(OH) "hydroxyapatite" and 9 mol of H₂O.

(b) What mass of hydroxyapatite could form from 100.0 g of 85% phosphoric acid and 100.0 g of calcium hydroxide?

  • Firstly, we need to calculate the no. of moles of 100.0 g of 85% phosphoric acid and 100.0 g of calcium hydroxide:

The no. of moles of 100.0 g of 85% phosphoric acid:

∵ The percent of phosphoric acid = 85%.

∴ The actual mass of phosphoric acid in solution = 85.0 g.

∴ no. of moles of phosphoric acid = mass/molar mass = (85.0 g)/(97.994 g/mol) = 0.87 mol.

The no. of moles of 100.0 g of calcium hydroxide:

∴ no. of moles of calcium hydroxide = mass/molar mass = (100.0 g)/(74.093 g/mol) = 1.35 mol.

From the stichiometry, Ca(OH)₂ react with H₃PO₄ with (5: 3) molar ratio.

∴ 1.35 mol of calcium hydroxide "limiting reactant" react completely with 0.813 mol of phosphoric acid "excess reactant" with (5: 3) molar ratio.

  • Now, we can find the no. of moles of produced Ca₅(PO₄)₃(OH) "hydroxyapatite":

Using cross multiplication:

5 mol of Ca(OH)₂ produce → 1 mol Ca₅(PO₄)₃(OH), from stichiometry.

1.35 mol of Ca(OH)₂ produce → ??? mol Ca₅(PO₄)₃(OH).

∴ The no. of moles of produced Ca₅(PO₄)₃(OH) "hydroxyapatite" = (1.35 mol)(1 mol)/(5 mol) = 0.27 mol.

Finally, we can get the mass of produced Ca₅(PO₄)₃(OH) "hydroxyapatite" = (no. of moles)(molar mass) = (0.27 mol)(502.3 g/mol) = 135.62 g.

mickymike92

The equation of the reaction is:

  • 5Ca(OH)₂ + 3H₃PO₄ ---> Ca₅(PO₄)₃(OH) + 9H₂O.

Based on the data provided, the mass of hydroxyapatite formed is  135.62 g.

What are minerals?

Minerals are substances which contain metallic elements combined with other elements found in the earth's crust.

Example of a mineral is Hydroxyapatite, Ca5(PO4)3(OH).

The balanced equation for the  preparation of hydroxyapatite is given below:

  • 5Ca(OH)₂ + 3H₃PO₄ ---> Ca₅(PO₄)₃(OH) + 9H₂O,

that 5 mol of Ca(OH)₂ react with 3 mol of H₃PO₄ to produce 1 mol of Ca₅(PO₄)₃(OH) and 9 mol of H₂O.

The mass of hydroxyapatite prepared from 100.0 g of 85% phosphoric acid and 100.0 g of calcium hydroxide? is calculated as follows:

100.0 g of 85% phosphoric acid contains 100 * 85/100 g of  phosphoric acid = 85.0 g.

moles of phosphoric acid in 85 g = mass/molar mass

molar mass of phosphoric acid = 97.994 g/mol

moles of phosphoric acid in 85 g = 85.0 g/97.994 g/mol

moles of phosphoric acid in 85 g = 0.87 moles

moles of calcium hydroxide in 100.0 g = mass/molar mass

molar mass of calcium hydroxide = 74.093 g/mol

moles of calcium hydroxide in 100.0 g = 100.0 g/74.093 g/mol

moles of calcium hydroxide in 100.0 g = 1.35 moles

From the equation of the reaction, Ca(OH)₂ react with H₃PO₄ in a 5: 3 molar ratio.

5 mol of Ca(OH)₂ produce 1 mole Ca₅(PO₄)₃(OH)

1.35 mol of Ca(OH)₂ will  produce 1.35 * 1/5 moles of Ca₅(PO₄)₃(OH) = 0.27 moles

mass of 0.27 moles of hydroxyapatite = no. of moles * molar mass

molar mass of hydroxyapatite = 502.3 g/mol

mass of hydroxyapatite =  0.27 mol * 502.3 g/mol

mass of hydroxyapatite = 135.62 g

Therefore, the mass of hydroxyapatite produced is 135.62 g

Learn more about minerals at: https://brainly.com/question/89259

Otras preguntas