How much 0.7MKI solution in liters will completely precipitate the Pb2+ in 2.1L of 0.18MPb(NO3)2 solution?

Question

Question

Answer 1

Answer #1

no of moles of Pb(NO3)2 = molarity * volume in L

= 0.18*2.1 = 0.378moles

2KI(aq) + Pb(NO3)2(aq) -----------> PbI2(s) + 2KNO3(aq)

1 mole of Pb(NO3)2 react with 2 moles of KI

0.378 moles of Pb(NO3)2 react with = 2*0.378/1 = 0.756moles of KI

no of moles of KI = molarity * volume in L

0.756 = 0.7* volume in L

volume in L = 0.756/0.7 = 1.08L

volume of KI = 1.08L >>>>answer

Answer 2

Similar Homework Help Questions

2KI (aq) + Pb(NO3)2 (aq) + PbI, (s) + 2KNO3 (aq). How much 0.7 M KI...

2KI (aq) + Pb(NO3)2 (aq) + PbI, (s) + 2KNO3 (aq). How much 0.7 M KI solution in liters will completely precipitate the Pb2+ in 2.1 L of 0.18 M Pb(NO3), solution? Do not include units in your answer and round to two significant figures. Druiden bela
how much 0.1MNaOH solution will completely precipitate the Mg2+ in 0.8L of 0.10MMgCl2 solution?

how much 0.1MNaOH solution will completely precipitate the Mg2+ in 0.8L of 0.10MMgCl2 solution?
KCl (aq) + AgNO3(aq) → KNO3(aq) + Agus) How much 0.9 M KCl solution in liters...

KCl (aq) + AgNO3(aq) → KNO3(aq) + Agus) How much 0.9 M KCl solution in liters will completely precipitate the Agion in 0.9 L of 0.19 M AgNO, solution? Round to two significant figures, and do not include units in your answer.

A solution is 0.0045 M in both Pb2+ and Ca2+. Solid Na2SO4 is added to precipitate...

A solution is 0.0045 M in both Pb2+ and Ca2+. Solid Na2SO4 is added to precipitate the sulfates. What concentration of SO42‒ is needed to precipitate as much of the Pb2+ without precipitating the Ca2+? Ksp(PbSO4) = 1.6 × 10‒8, Ksp(CaSO4) = 2.4 × 10‒4
Refer to the precipitation reaction below. CoCl2(aq)+2NaOH(ag) Co(OH)2(s)+2NaCI(aq) How much 0.5 M NaOH solution in liters...

Refer to the precipitation reaction below. CoCl2(aq)+2NaOH(ag) Co(OH)2(s)+2NaCI(aq) How much 0.5 M NaOH solution in liters will completely precipitate the Co+ in 1.6 L of 0.12 M CoCl solution? Round to two significant figures, and do not include units in your answer Provide your answer below
If a solution containing 30.61 g of lead(II) nitrate is allowed to react completely with a...

If a solution containing 30.61 g of lead(II) nitrate is allowed to react completely with a solution containing 5.102 g of sodium sulfide, how many grams of solid precipitate will be formed? mass of solid precipitate: How many grams of the reactant in excess will remain after the reaction? mass of excess reactant: Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (O) for the number...

ASAP What volume, in liters, of 1.00 M Pb(NO3)2 is needed to react completely with 0.500...

ASAP What volume, in liters, of 1.00 M Pb(NO3)2 is needed to react completely with 0.500 L of 4.00 M NaCl, according to the following equation? Pb(NO3)2(aq) + 2 NaCl(ag) – PbCl2(s) + 2 NaNO3(aq) Upload Choose a File
2:06 LTED Question 10 of 20 Submit How many liters of a 0.209 MKI solution is needed to completely react with 2.43...

2:06 LTED Question 10 of 20 Submit How many liters of a 0.209 MKI solution is needed to completely react with 2.43 g of Cu(NO3)2 according to the balanced chemical reaction: 2 2 Cu(NO3)2(aq) + Cul(aq) + 12(s) + 4 4 Kl(aq) → KNO3(aq) STARTING AMOUNT ADD FACTOR ANSWER RESET 2.43 2 0.124 6.022 x 1023 187.57 0.00542 0.0310 0.0620 0.209 166 4 Tap here or pull up for additional resources
How many liters of 0.122 M NaOH solution are needed to react completely with 0.300 L...

How many liters of 0.122 M NaOH solution are needed to react completely with 0.300 L of 0.672 M H2SO4 solution? 2 NaOH + H2SO4 2 H2O + Na2SO4

How many grams Ba(NO3)2 of are required to precipitate all the sulfate ion present in 15.3...

How many grams Ba(NO3)2 of are required to precipitate all the sulfate ion present in 15.3 mL of 0.143 M H2SO4 solution? Ba(NO3)2 + H2SO4 --> BaSO4 + 2HNO3