Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO,),(aq) + 2 NH,I(aq) PbI,(s) + 2 NH,NO,(aq) What volume of a 0.290 M NHẠI solution is required to react with 447 mL of a 0.320 M Pb(NO3)2 solution? volume: mL How many moles of PbI2 are formed from this reaction? moles: mol PbI2

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Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction
Pb(NO,),(aq) + 2 NH,I(aq)
PbI, (s) + 2 NH,NO3 (aq)
What volume of a 0.290 M NHẠI solution is required to react with 447 mL of a 0.320 M Pb(NO3)2 solution?
volume:
mL
How many moles of PbI2 are formed from this reaction?
moles:
mol PbI2
Transcribed Image Text:Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO,),(aq) + 2 NH,I(aq) PbI, (s) + 2 NH,NO3 (aq) What volume of a 0.290 M NHẠI solution is required to react with 447 mL of a 0.320 M Pb(NO3)2 solution? volume: mL How many moles of PbI2 are formed from this reaction? moles: mol PbI2
Expert Solution
Step 1

Given the volume of 0.320 M Pb(NO3)2(aq) taken = 447 mL * (1L/1000 mL) = 0.447 L

=> moles of Pb(NO3)2(aq) taken = 0.320 mol/L * 0.447 L = 0.14304 mol

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