To determine the limiting reactant, we need to calculate the amount of product that would be produced from each reactant separately, and then compare those values to see which reactant would produce the lower amount of product, since that would be the limiting reactant.
First, we need to convert the masses of reactants given in the problem to moles.
7.000 g NaCl x (1 mol NaCl/22.990 g NaCl) = 0.304 mol NaCl
50.400 g Pb(NO3)2 x (1 mol Pb(NO3)2/331.200 g Pb(NO3)2) = 0.152 mol Pb(NO3)2
Next, we need to use stoichiometry to calculate the amount of product (PbCl2) that would be produced from each reactant. From the balanced chemical equation, we see that 2 moles of NaCl react with 1 mole of Pb(NO3)2 to produce 1 mole of PbCl2.
0.304 mol NaCl x (1 mol PbCl2/2 mol NaCl) = 0.152 mol PbCl2
0.152 mol Pb(NO3)2 x (1 mol PbCl2/1 mol Pb(NO3)2) = 0.152 mol PbCl2
We can see from these calculations that the amount of PbCl2 that would be produced from Pb(NO3)2 is equal to the amount that would be produced from NaCl. Therefore, NaCl is the limiting reactant since it would be completely used up before all of the Pb(NO3)2 has a chance to react.
Answer: NaCl is the limiting reactant.
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
NaNO3
NaCl
Pb(NO3)2
PbCl2
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