Question

For the reaction 1Pb(NO3)2 + 2KI −−> 1PbI2 + 2KNO3, how many moles of lead iodide (PbI2) are produced from 300.00 g of potassium iodide (KI)?

Answer 1

moles PbI₂ produced = = 0.9 mole (1 sig. fig.)

Step-by-step explanation:

Given Pb(NO₃)₂ + 2KI −−> PbI₂ + 2KNO₃

300g ?moles

moles KI = 300g KI/166g/mol = 1.81 mole KI

Rxn ratio for KI and PbI₂ => 2:1

∴ moles PbI₂ produced = 1/2(1.81 mole) = 0.9036144 mole (calculator answer)

= 0.9 mole (1 sig. fig.)