When a solution of lead nitrate, Pb(NO_3)_2, is mixed with a solution of sodium

Question

When a solution of lead nitrate, Pb(NO_3)_2, is mixed with a solution of sodium iodide, Nal, a precipitate of lead iodide, PbI_2, is formed. (1) Write the chemical equation for this reaction showing the state of all reactants and products. (2) If one mole of Pb(NO_3)_2 reacts with two moles of Nal, how many moles of Pbl_2 will be formed? (3) If two moles of Pb(NO_3)_2 reacts with two moles of Nal, which compound would be the limiting reagent? (4) How many moles of PbI_2 will be formed in (3) above?

Explanation / Answer

(a) Balanced equation is,

Pb(NO3)2 (aq.) + 2 NaI (aq.) ------------> PbI2 (aq.) + 2 NaNO3 (aq.)

(b)

From the balanced equation,

1 mol Pb(NO3)2 reacts with 2 mol of NaI to form 1 mol of PbI2

(c)

From the balanced equation for every 1 mol of Pb(NO3)2, 2 mol of NaI is required.

Hence, for 2 mol of Pb(NO3)2, 4 mol of NaI is required. (but only 2 mol is there)

Hence, NaI is limiting reqgent.

(d)

From the balanced equation,

2 mol of NaI can form 1 mol of PbI2.

Hence moles of PbI2 formed = 1 mol

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