Answer :

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Taking into account the reaction stoichiometry, 112.61 grams of lead(II) iodide, PbI₂, are produced from 81.1 g of potassium iodide, KI.

Reaction stoichiometry

In first place, the balanced reaction is:

2 KI + Pb(NO₃)₂  → PbI₂+ 2 KNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • KI: 2 moles
  • Pb(NO₃)₂: 1 mole
  • PbI₂: 1 mole
  • KNO₃: 2 moles

The molar mass of the compounds is:

  • KI: 166 g/mole
  • Pb(NO₃)₂: 331.2 g/mole
  • PbI₂: 461 g/mole
  • KNO₃: 101.1 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • KI: 2 moles ×166 g/mole= 332 grams
  • Pb(NO₃)₂: 1 mole ×331.2 g/mole= 331.2 grams
  • PbI₂: 1 mole ×461 g/mole= 461 grams
  • KNO₃: 2 moles ×101.1 g/mole= 202.2 grams

Mass of PbI₂ formed

The following rule of three can be applied: if by reaction stoichiometry 332 grams of KI form 461 grams of PbI₂, 81.1 grams of KI form how much mass of PbI₂?

mass of PbI₂= (81.1 grams of KI× 461 grams of PbI₂)÷ 332 grams of KI

mass of PbI₂= 112.61 grams

Then, 112.61 grams of lead(II) iodide, PbI₂, are produced from 81.1 g of potassium iodide, KI.

Learn more about the reaction stoichiometry:

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