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 oxygen atoms are there in 1.75 ng of Ca3(PO4)2?
A)3.4 * 10^12
B)2.72 * 10^13

Answer 1

The number of oxygen atoms in 1.75 ng of Ca3(PO4)2 is 4.512 x 10^-8.

Step-by-step explanation:

To find the number of oxygen atoms in 1.75 ng of Ca3(PO4)2, we first need to determine the molar mass of Ca3(PO4)2. The molar mass can be calculated by adding up the atomic masses of all the atoms in the compound. In Ca3(PO4)2, there are 3 calcium atoms, 2 phosphorus atoms, and 8 oxygen atoms.

The molar mass of Ca3(PO4)2 can be calculated as follows:

(3 x atomic mass of calcium) + (2 x atomic mass of phosphorus) + (8 x atomic mass of oxygen)
= (3 x 40.08 g/mol) + (2 x 30.97 g/mol) + (8 x 16.00 g/mol)
= 120.24 g/mol + 61.94 g/mol + 128.00 g/mol
= 310.18 g/mol

Now, we can calculate the number of moles of Ca3(PO4)2:

Number of moles = mass in grams / molar mass
= 1.75 ng / (310.18 g/mol x 1,000,000)
= 5.64 x 10^-9 mol

Finally, we can calculate the number of oxygen atoms in 1.75 ng of Ca3(PO4)2:

Number of oxygen atoms = 5.64 x 10^-9 mol x 8
= 4.512 x 10^-8