To determine the limiting reactant, we need to calculate the amount of product that can be produced from each reactant and identify the reactant that produces the lowest amount of product.
First, we need to calculate the number of moles of each reactant:
Number of moles of NaCl = 7.000 g / 22.990 g/mol = 0.304 mol
Number of moles of Pb(NO3)2 = 50.400 g / 331.200 g/mol = 0.152 mol
Next, we need to determine the limiting reactant by calculating the amount of product that can be produced from each reactant:
From the balanced chemical equation, we know that 2 moles of NaCl react with 1 mole of Pb(NO3)2 to produce 1 mole of PbCl2. Therefore, the theoretical amount of PbCl2 that can be produced is:
Theoretical amount of PbCl2 from NaCl = 0.304 mol NaCl x (1 mol PbCl2 / 2 mol NaCl) = 0.152 mol PbCl2
Theoretical amount of PbCl2 from Pb(NO3)2 = 0.152 mol Pb(NO3)2 x (1 mol PbCl2 / 1 mol Pb(NO3)2) = 0.152 mol PbCl2
Based on these calculations, we see that both reactants produce the same theoretical amount of PbCl2. Therefore, neither reactant is in excess and both are limiting reactants.
In this case, we could have predicted that NaCl would be in excess since it has a higher amount of moles than Pb(NO3)2, but we need to compare the theoretical amounts of product produced to determine the actual limiting reactant.
The actual amount of PbCl2 that can be produced will be limited by the reactant that produces the lowest amount of product, which in this case is 0.152 mol from both reactants. Therefore, the limiting reactant is both NaCl and Pb(NO3)2.
The amount of lead (II) chloride that can be produced from the reaction is 0.152 mol, which is equivalent to:
0.152 mol x 207.200 g/mol = 31.462 g PbCl2.
Lead (II) chloride (PbCl2) are produced from the reaction of 7.000 g of NaCl with 50.400 g of Pb(NO3)2? What is the limiting reactant? (Atomic mass: Na-22.990 g/mol; 0-15.999 g/mol; Cl-35.453 g/mol; Pb-207.200 g/mol; N-14.007).
2NaCl+ Pb(NO3)2 → 2 NaNO3 + PbCl2
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