ge-CHEM 4-Summer18 . HAN > Activities and Due Dates) HW 16-2 8/5/2018 11:55 PM 2
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ge-CHEM 4-Summer18 . HAN > Activities and Due Dates) HW 16-2 8/5/2018 11:55 PM 2.1/5 8/4/2018 03:08 PM 3 Gradebook Print Calculator Periodic Table Question 10 of 16 Sapling Learning Map Lead(ll) nitrate and ammonium iodide react to form lead(ll) iodide and ammonium nitrate according to the reaction What volume of a 0.110 M NHl solution is required to react with 263 mL of a 0.100 M P(NOsh2 solution? Number mL How many moles of Pbl2 are formed from this reaction? Number mol Pb, Previous ?Give Up & View Scubon e' ck Answer ONext Hint Eut about us careers privacy policy tes of use contact us heipExplanation / Answer
Given reaction
Pb(NO3)2+2NH4I------->PbI2+2NH4NO3
Here 2 moles of ammonium iodide reacts with 1 mole of lead nitrate.
Molarity of lead nitrate=0.100M
volume of lead nitrate=263 ml
total number of moles of lead nitrate=Molarity *volume
=26.3 millimoles
therefore number of moles of ammonium iodide required=26.3*2 millimoles=52.6 millimoles
Molarity of ammonium iodide=0.110M
Molarity=number of moles/volume
therefore volume required=number of moles/molarity=52.6/0.110=478.18ml
Also for each mole of lead nitrate,one mole of lead iodide is formed
Therefore number of lead Iodide formed =26.3 millimoles
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