In the Reaction: Pb(NO3)2 + 2 Kl —> Pbl2 + 2 KNO3 How many grams of lead (II) iodide, Pbl2, is produced from 82.7 grams of potassium iodide, KI?

Round to two decimal places.

114.8g PbI2 are produced

Explanation:

To solve this question we must convert the mass of KI to moles using its molar mass (KI = 166.0g/mol). As 2 moles of KI produce 1 mole of PbI2 we can find its moles and its mass (Molar mass PbI2: 461.01g/mol) as follows:

Moles KI:

82.7g KI * (1mol / 166.0g) = 0.498 moles KI

Moles PbI2:

0.498 moles KI * (1mol PbI2 / 2mol KI) = 0.249 moles PbI2

Mass PbI2:

0.249moles * (461.01g/mol) =

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

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