Question

If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + NaI → PbI2 + NaNO3

Answer 1

Answer:- 27.7 grams of $PbI_2$ are produced.

Solution:- The balanced equation is:

$Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3$

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.

Molar mass of $Pb(NO_3)_2 = 207.2+2(14.01)+6(16) = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2$ = 207.2+2(126.90) = 461 gram per mol

let's do the calculations for the grams of the product for the given grams of each of the reactant:

$28.0gPb(NO_3)_2(\frac{1molPb(NO_3)_2}{331.22gPb(NO_3)_2})(\frac{1molPbI_2}{1molPb(NO_3)_2})(\frac{461gPbI_2}{1molPbI_2})$

= $39.0gPbI_2$

$18.0gNaI(\frac{1molNaI}{149.89gNaI})(\frac{1molPbI_2}{2molNaI})(\frac{461gPbI_2}{1molPbI_2})$

= $27.7gPbI_2$

From above calculations, NaI gives least amount of $PbI_2$, so the answer is, 27.7 g of $PbI_2$ are produced.