Question

. How many GRAMS of lead(II) iodide can be made from 2.00 grams of lead(II) ?

Answer 1

first write balanced equation

Pb²⁺ + 2I^- ----> PbI₂

from the balanced equation it is clear that on emole of lead(II) will give on emole of lead(II) iodide

first find out the no of moles of Pb2+ = weight of Pb2+ / molar mass of Pb2+

remember molar mass of Pb = remember molass of Pb2+

= 2.0 g / 207.2 g/mole

= 0.0096 moles

from the balanced equation

0.0096 moles of Pb2+ will give 0.0096 moles of PbI2

so moles of PbI2 = 0.0096

now use the formula

moles = weight/molar mass

weight of PbI₂ = moles of PbI₂ x molar mass of PbI₂

= 0.0096 moles x 461.01 g/mol

= 4.45 grams PbI2