Question

How many GRAMS of lead(II) are present in 3.13 grams of lead(II) iodide ?

Answer 1

lead(II) iodide molecular weight = 461.01

we have 3.13 gm of lead(II) iodide

converts into moles

number of moles = wt/M.wt = 3.13/461.01 = 0.006789 moles

lead(II) iodide (PbI₂) contains one Pb and two iodine atoms. So in 1 mole of PbI₂ contains one mole of Pb and two moles of iodine.

So in 0.006789 moles of PbI₂ contains 0.006789 moles of Pb atoms.

Concerts moles into grams

Pb molecular weight = 207.98

Moles = weight (in gm)/M.Wt

0.006789 = wt/207.98

Wt=1.411 gm

3.13 grams of lead(II) iodide contains 1.411 gm of led