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
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.
To determine the limiting reactant, we need to compare the amount of product that can be formed from each reactant. The reactant that produces the smaller amount of product is the limiting reactant.
1. Find the molar masses of NaCl and Pb(NO3)2:
Molar mass of NaCl = 22.990 g/mol (Na) + 35.453 g/mol (Cl) = 58.443 g/mol
Molar mass of Pb(NO3)2 = 207.200 g/mol (Pb) + 2 * (14.007 g/mol (N) + 3 * 16.000 g/mol (O)) = 331.200 g/mol
2. Convert the mass of each reactant to moles:
Moles of NaCl = 7.000 g / 58.443 g/mol ≈ 0.120 mol
Moles of Pb(NO3)2 = 50.400 g / 331.200 g/mol ≈ 0.152 mol
3. Use the stoichiometry of the balanced equation to determine the moles of PbCl2 that can be formed from each reactant:
From the balanced equation, we can see that 2 moles of NaCl react with 1 mole of Pb(NO3)2 to produce 1 mole of PbCl2.
Moles of PbCl2 from NaCl = 0.120 mol NaCl * (1 mol PbCl2 / 2 mol NaCl) = 0.060 mol
Moles of PbCl2 from Pb(NO3)2 = 0.152 mol Pb(NO3)2 * (1 mol PbCl2 / 1 mol Pb(NO3)2) = 0.152 mol
4. Compare the moles of PbCl2 formed from each reactant:
Moles of PbCl2 formed from NaCl: 0.060 mol
Moles of PbCl2 formed from Pb(NO3)2: 0.152 mol
Since 0.060 mol < 0.152 mol, the limiting reactant is NaCl.
To determine the limiting reactant, we need to compare the amounts of products that can be formed from each reactant.
First, let's calculate the moles of each reactant:
For NaCl:
Mass of NaCl = 7.000 g
Molar mass of NaCl = 22.990 g/mol + 35.453 g/mol = 58.443 g/mol
Moles of NaCl = 7.000 g / 58.443 g/mol ≈ 0.120 mol
For Pb(NO3)2:
Mass of Pb(NO3)2 = 50.400 g
Molar mass of Pb(NO3)2 = 207.200 g/mol + 2 * (14.007 g/mol + 15.999 g/mol × 3) = 331.202 g/mol
Moles of Pb(NO3)2 = 50.400 g / 331.202 g/mol ≈ 0.152 mol
Next, we need to compare the stoichiometric ratio of the reactants to the products to determine which reactant will limit the amount of product formed.
From the balanced equation:
2 moles of NaCl react with 1 mole of Pb(NO3)2 to produce 1 mole of PbCl2
Comparing the two moles of reactants, we can see that the ratio of NaCl to Pb(NO3)2 is 2:1.
Now, let's determine how many moles of PbCl2 can be formed from each reactant.
For NaCl:
Moles of PbCl2 formed from NaCl = 0.120 mol × (1 mol PbCl2 / 2 mol NaCl) = 0.060 mol
For Pb(NO3)2:
Moles of PbCl2 formed from Pb(NO3)2 = 0.152 mol × (1 mol PbCl2 / 1 mol Pb(NO3)2) = 0.152 mol
From the calculations, we can see that the moles of PbCl2 formed from NaCl and Pb(NO3)2 are 0.060 mol and 0.152 mol, respectively.
The limiting reactant is the one that produces the lesser amount of product. In this case, NaCl produces 0.060 mol of PbCl2, while Pb(NO3)2 produces 0.152 mol of PbCl2.
Therefore, the limiting reactant is NaCl.