Answered

Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

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 :

drpelezo

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.)

Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.