Question

Phosphoric acid can be prepared by reaction of sulfuric acid with “phosphate rock” according to the equation: Ca3(PO4)2 3H2SO4 → 3CaSO4 2H3PO4 How many formula units of Ca3(PO4)2 are there in 4.6E-9 g

asked Feb 6, 2024 118k views

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Uday Hiwarale · Answer 1 · 2024-02-11T08:10:22+0000

We calculated that there are 8.93E12 formula units of Ca3(PO4)2 in 4.6E-9 g of Ca3(PO4)2.

Number of formula units of Ca3(PO4)2 in 4.6E-9 g of Ca3(PO4)2

To calculate the number of formula units of Ca3(PO4)2 in 4.6E-9 g of Ca3(PO4)2, we can follow these steps:

Calculate the molar mass of Ca3(PO4)2. The molar mass of Ca3(PO4)2 is the sum of the atomic masses of each atom in the formula unit.

Molar mass of Ca3(PO4)2

= 3 * Atomic mass of Ca

+ 2 * Atomic mass of P

+ 8 * Atomic mass of O

= 3 * 40.08 g/mol

+ 2 * 30.97 g/mol

+ 8 * 15.9994 g/mol

= 310.18 g/mol

Convert the mass of Ca3(PO4)2 to moles. We can use the following formula to convert the mass (m) of a substance to moles (n):

n = m / M

where:

n is the number of moles

m is the mass of the substance

M is the molar mass of the substance

In this case, we have:

n(Ca3(PO4)2) = 4.6E-9 g / 310.18 g/mol = 1.48E-11 mol

Calculate the number of formula units. The number of formula units is equal to the number of moles times the Avogadro's constant (6.022E23 mol^-1).

Number of formula units = n(Ca3(PO4)2) * Avogadro's constant

= 1.48E-11 mol * 6.022E23 mol^-1

= 8.93E12

Therefore, there are 8.93E12 formula units of Ca3(PO4)2 in 4.6E-9 g of Ca3(PO4)2.