3/2312018 09:00 AM 4 11.8/20 33120/18 1:20 AM Gradbook Print Calculator -A Perio

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3/2312018 09:00 AM 4 11.8/20 33120/18 1:20 AM Gradbook Print Calculator -A Periodic Table Question 8 of 10 Sapling Learning Map . Lead(II) nitrate and ammonium iodide react to form lead(lI) iodide and ammonium nitrate according to the reaction Pb(NO3.(aq) +2NH3(aq) 11s) + 2 NH,NO,(aq) What volume of a 0.530 M NH&l; solution is required to react with 789 mL of a 0.120 M Pb(NOs)2 solution? Number mL How many moles of Pbl2 are formed from this reaction? Number mol Pbl, & Hint O Previous Give Up & View Solution Q Check Answer 0 Next Exit about us careers privacy policy terms of use contact us help

Explanation / Answer

Number of moles of Pb(NO3)2 = moalrity * volume of solution in L

Number of moles of Pb(NO3)2 = 0.120 * 0.789 = 0.0947 mole

from the balanced equation we can say that

1 mole of Pb(NO3)2 contains 2 mole of NH4I

so 0.0947 mole of  Pb(NO3)2 will contain

= 0.0947 mole of  Pb(NO3)2 *(2 mole of NH4I / 1 mole of Pb(NO3)2 )

= 0.189 mole of NH4I

Molarity = number of moles of NH4I / volume of solution in L

0.530 = 0.189 mole / volume of solution in L

volume of solution in L = 0.189 / 0.530 = 0.357 L

1 L = 1000 mL

0.357 L = 357 mL

Therefore, the volume of NH4I = 357 mL

1 mole of Pb(NO3)2 produces 1 mole of PbI2 so

0.0947 mole of Pb(NO3)2 will produce 0.0947 mole of PbI2

Therefore, the number moles of PbI2 = 0.0947

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