Reaction between lead(ii) nitrate and sodium iodide to produce lead(ii) iodide and sodium nitrate

Md Ariful Islam

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Md Ariful Islam

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Pb(NO3)2 (aq) + 2 NaI (aq) → PbI2 (s) + 2 NaNO3 (aq)

2Na(+) +2I(-) + Pb(2+) + 2NO3(-) = 2Na(+) +PbI2(s) + 2NO3(-)

2I(-) + Pb(2+) = PbI2 (s)

Lead iodide appears as a yellow crystalline solid. Insoluble in water.

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For the reaction Pb(NO<sub>3</sub>)<sub>2</sub> + 2KI → PbI<sub>2</sub> + 2KNO<sub>3</sub>, how many moles of lead iodide are produced from 350. g of potassium iodide?

You13:171) n(KI)= m(KI)/M(KI)= 350/166=2.108 mole2) n(PI2)=n(KI)/2=2.108/2= 1.05 moleAnswer: 1.05 mole PbI2

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lead-204 203.973amu, 1.40% abundance. lead-206 205.974 amu, 24.10% abundance. lead-207???, abundance 22.10%. lead-208 207.977, abundance 52.40%. Calculate the isotope mass of lead-207

=207.2 amu (Pb)

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4gm of lead nitrate reacts with 1.17g of sodium chloride and form 3gm of lead chloride how much sodium nitrate will form?

Sodium Nitrate =4.73g

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Which one is more reactive? A- Methyl iodide B- Ethyl iodide C- Propyl iodide D- Isopropyl iodide Please address the query as soon as possible. Thank You!

B- Ethyl iodide

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write balanced chemical equations for each, remember to make your molecules neutral first. Sodium iodide and potassium hydroxide form when aqueous iodide is mixed with sodium hydroxide solution.

KI+NaOH=NaI+KOH

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If you start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?

Pb(NO3)2 + 2NaI = 2NaNO3 + PbI2n = m / MM (Pb(NO3)2) = 331.2 g/molM (NaI) = 150 g/molM (NaNO3) = 85 g/moln (Pb(NO3)2) = 25 / 331.2 = 0.08...

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What is the balanced equation to a solution of nitrate is combined with a solution of sodium carbonate to produce sodium nitrate solution and solid silver carbonate

Na2CO3 + 2AgNO3 = Ag2CO3 + 2NaNO3

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A 0.4230 g sample of impure sodium nitrate was heated, converting all the sodium nitrate to .1235 g of sodium nitrate and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Answer:2HNO3 = 2HNO2 + O2m= 85 x 0.1235 / 69= 0.1521W%= 0.1521 / 0.4230 = 36%

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Which of the above equilibria is forced to shift furthest to the left? Explain. - Lead nitrate and Sodium chloride - Lead chloride and Sodium chromate -Lead chromate and Sodium sulphide

Answer: Combination which makes less soluble product shift towards left side of the equation.Pb(NO3)2 + 2NaCl = PbCl2 + 2NaNO3PbCl2 + Na2CrO4 = PbCrO4 + 2NaClPbCrO4 + Na2S = PbS...

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When lead nitrate reacts with sodium iodide, sodium nitrate and lead iodide are formed.

This is the balanced equation of the reaction Pb(NO3)2 +2NaI → 2NaNO3 + PbI2

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