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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)?

Sagot :

Drpelezoidn

Answer:

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

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.)

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