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a Amazon.com: Teenwol! × New Tab xidentify acids and bases x Naming Worksheets gleaming corn/ibiscms/mod/ibisiew php?id=4926557 ng ate University-CHEM 1341 Spring18-PATTERSON (Sec. 257,259, and 262) Activities and Due Dates Chapter 4 Homework 3/1/2018 08:00 AM 26.8/100 , Prrt Calculator n Periodic Table Period. Table Question 31 of 33 Sapling Learning Map Lead(ll) nitrate and ammonium iodide react to form lead(ll) lodide and ammonium nitrate according to the reaction What volume of a 0.710 M NHal solution is required to react with 621 mL of a 0.320 M Pb(NOsl2 solution? Number 2755.68 mL How many moles of Pbl2 are formed from this reaction? Number 4409 mol Pbl, There is a hint available View the hint boniom drvider divider bar again to hide the hint. on the on the Close Previous Give Up & View Solton Check Answer ONet Exit Hint about uscamprivacy policy of usecontactuhelp earch

Explanation / Answer

The reaction is Pb(NO3)2 (aq)+2NH4I(aq) -------->PbI2(s)+2NH4NO3(s)

the stoichiometry suggests that 1 mole of Pb(NO3)2 reacts with 2 mole of NH4I to produce 1 mole of PbI2.

moles of Pb(NO3)2= molarity* Volume in Liters= 0.320*621/1000 =0.19872 moles

moles of NH4I =2 * moles of Pb(NO3)2= 2*0.19872= 0.39744

hence volume of NH4I in Liters = moles/Molarity=0.39744/0.710=0.560L or 0.560L*1000ml/L= 560 ml

2. moles of PbI2 fomed= moles of Pb(NO3)2 used =0.19872

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