if 1.37 moles of Pb(NO3)2 is mixed with 2.56 moles of KI, how many moles of PbI2 are produced?
Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3
Convert each to mols product. The smaller value will be the mols PbI2 produced.
1.37 mols Pb(NO3)2 x [1 mol PbI2/1 mol Pb(NO3)2] = 1.37 mols PbI2 produced.
2.56 mols KI x (1 mol PbI2/2 mols KI) = 1.28 mols PbI2 produced.
This is a limiting reagent problem. Obviously you may produced ONLY the smaller amount.
To determine the number of moles of PbI2 produced, we need to use the stoichiometry of the balanced chemical equation between Pb(NO3)2 and KI.
The balanced equation is:
Pb(NO3)2 + 2KI → PbI2 + 2KNO3
From the balanced equation, we can see that 1 mole of Pb(NO3)2 reacts with 2 moles of KI to produce 1 mole of PbI2.
Given:
- Moles of Pb(NO3)2 = 1.37 moles
- Moles of KI = 2.56 moles
Since the stoichiometric ratio is 1:2 between Pb(NO3)2 and KI, we need to determine the limiting reactant. The limiting reactant is the one that provides fewer moles than required by the stoichiometry.
To find the limiting reactant, we compare the moles of each reactant with their stoichiometric coefficients:
Moles of Pb(NO3)2 / Stoichiometric coefficient of Pb(NO3)2 = 1.37 / 1 = 1.37
Moles of KI / Stoichiometric coefficient of KI = 2.56 / 2 = 1.28
Comparing the values, we can see that KI has a smaller value. Therefore, KI is the limiting reactant.
Now, we can use the stoichiometry to determine the moles of PbI2 produced. Since 2 moles of KI react with 1 mole of PbI2, we can calculate it as follows:
Moles of KI (limiting reactant) * (1 mole of PbI2 / 2 moles of KI) = 1.28 * (1/2) = 0.64 moles
Hence, when 1.37 moles of Pb(NO3)2 is mixed with 2.56 moles of KI, 0.64 moles of PbI2 are produced.