Question

A 3.00 L sample of paint that has a density of 4.65 g/mL is found to contain 33.1 g lead (II) nitride. How many ions of lead are in the paint?

Answer 1

We are asked to calculate the number of ions of lead in paint.

Given that paint has:

Volume = 3.00 L

density = 4.65 g/mL

The mass of Lead (II) nitride in paint is = 33.1 g

Firstly let us write the equation:

`Pb_3N_2\text{ }\rightarrow3Pb^(2+)+2N^(3-)`

We can calculate the number of moles of Pb3N2 and use stoichiometry to find the number of moles of Pb.

number of moles (n) of Pb3N2:

n = m/M where m is the mass and M is the molar mass of Pb3N2

n = 33.1g/649.61 g/mol

n = 0.0509 moles

number of moles of Pb:

Using the stoichiometry, the molar ratio between Pb3N2 and Pb is 1:3 according to the equation.

Therefore number of moles of Pb is 0.0509 moles x 3 = 0.1529 moles.

After calculating the number of moles, the number of ions will be equal to the product of the number of moles and Avogadro's number.

number of Pb ions = number of moles x avogadros number

number of Pb ions = 0.1529 moles x 6.022x10^23

number of Pb ions = 9.2076x10^22