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. Determine (A) how many grams of lead ion are in the paint sample? (B) How many ions of lead are in the paint? (C) what is mass percentage of lead in the paint/ What is the ppm?

Answer 1

1) Grams of lead.

Convert grams of lead (II) nitride to moles of lead (II) nitride

`molofPb_3N_2=33.1gPb_3N_2\cdot(1molPb_3N_2)/(649.61gPb_3N_2)=0.0509lofPb_3N_2`

Convert moles of lead (II) nitride into moles of Pb

`\text{molesofPb}=0.0509molPb_3N_2\cdot\frac{3\text{molPb}}{1molPb_3N_2}=0.1527\text{molPb}`

Convert moles of Pb into grams of Pb

`\text{gofPb}=0.1527\text{molPb}\cdot\frac{207.2\text{ g Pb}}{1\text{molPb}}=31.64\text{ g Pb}`

There is 31.64 g of ions of lead in the sample of paint.