Answer: 50.7 grams
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
a) moles of [tex]KI[/tex]
[tex]\text{Number of moles}=molarity\times {\text {Volume in L]}=0.417M\times 0.528L=0.220moles[/tex]
The balanced chemical equation is:
[tex]2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)[/tex]
[tex]KI[/tex] is the limiting reagent as it limits the formation of product and [tex]Pb(NO_3)_2[/tex] is in excess.
According to stoichiometry :
2 moles of [tex]KI[/tex] give = 1 mole of [tex]PbI_2[/tex]
Thus 0.220 moles of [tex]KI[/tex] give=[tex]\frac{1}{2}\times 0.220=0.110moles[/tex] of [tex]PbI_2[/tex]
Mass of [tex]PbI_2=moles\times {\text {Molar mass}}=0.110moles\times 461.01g/mol=50.7g[/tex]
Thus 50.7 g of [tex]PbI_2[/tex] will be formed.
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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