Consider the balanced equation of k i ki reacting with p b ( n o 3 ) 2 pb(nox3)x2 to form a precipitate. 2 k i ( a q ) + p b ( n o 3 ) 2 ( a q ) ⟶ p b i 2 ( s ) + 2 k n o 3 ( a q ) 2ki(aq)+pb(nox3)x2(aq)⟶pbix2(s)+2kn ox3(aq) what mass of p b i 2 pbix2 can be formed by adding 0.528 l of a 0.417 m solution of k i ki to a solution of excess p b ( n o 3 ) 2 pb(nox3)x2?

50.7 g of precipitate is produced.

The equation of the reaction is; 2KI(aq) + Pb(NO3)2(aq) ⟶ PbI2(s) + 2KNO3 (aq)

We already know from the information in the question that the reactant in excess is Pb(NO3)2.

Number of moles of PbI2= 0.417 M × 0.528 L = 0.22 moles

From the stoichiometry of the reaction;

2 moles of KI yields 1 mole of PbI2

0.22 moles of KI yields  0.22 moles × 1 mole/2 moles

= 0.11 moles of PbI2

Mass of PbI2produced = number of moles of PbI2×  molar mass of PbI2

Molar mass of PbI2= 461 g/mol

Mass of PbI2produced = 0.11 moles of PbI2× 461 g/mol = 50.7 g of PbI2

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50.7 grams

Explanation:

To calculate the moles, we use the equation:

a) moles of

The balanced chemical equation is:

is the limiting reagent as it limits the formation of product and is in excess.

According to stoichiometry :

2 moles of give =  1 mole of

Thus 0.220 moles of give=  of

Mass of

Thus 50.7 g of will be formed.

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