How many grams of lead (II) chloride (PbCl2) are produced from the reaction of 7.000 g of NaCl and 50.400 g of Pb(NO3)2?Calculate the theoritical yield of PbCl2? (Atomic mass: Na-22.990 g/mol; 0-15.999 g/mol; Cl-35.453g/mol; Pb-207.200 g/mol; N-14.007). Hint: Determine the limiting reagent first before calculating the theoritical yield of PbCl2.
2NaCl+ Pb(NO3)2 → 2 NaNO3 + PbCl2
33.373 g
42.272 g
16.547 g
21.136 g
To determine the limiting reagent, we need to compare the amount of product that can be produced from each reactant. Let's start with NaCl:
From the balanced equation, 2 moles of NaCl reacts with 1 mole of Pb(NO3)2 to produce 1 mole of PbCl2. Since we have 7.000 g of NaCl, we need to convert it to moles:
7.000 g NaCl x (1 mol NaCl/22.990 g NaCl) = 0.304 mol NaCl
Since the mole ratio of NaCl to PbCl2 is 2:1, we can calculate the theoretical yield of PbCl2 that can be produced from 0.304 mol NaCl:
0.304 mol NaCl x (1 mol PbCl2/2 mol NaCl) x (207.200 g PbCl2/1 mol PbCl2) = 31.66 g PbCl2
Now let's do the same calculations for Pb(NO3)2:
From the balanced equation, 1 mole of Pb(NO3)2 reacts with 2 moles of NaCl to produce 1 mole of PbCl2. Since we have 50.400 g of Pb(NO3)2, we need to convert it to moles:
50.400 g Pb(NO3)2 x (1 mol Pb(NO3)2/331.200 g Pb(NO3)2) = 0.152 mol Pb(NO3)2
Since the mole ratio of Pb(NO3)2 to PbCl2 is 1:1, we can calculate the theoretical yield of PbCl2 that can be produced from 0.152 mol Pb(NO3)2:
0.152 mol Pb(NO3)2 x (1 mol PbCl2/1 mol Pb(NO3)2) x (207.200 g PbCl2/1 mol PbCl2) = 31.99 g PbCl2
Since the theoretical yield of PbCl2 from NaCl is lower than from Pb(NO3)2, NaCl is the limiting reagent. Therefore, the maximum amount of PbCl2 that can be produced is 31.66 g.
Answer: 31.66 g PbCl2
To determine the limiting reagent, we need to compare the number of moles of NaCl and Pb(NO3)2.
1. Calculate the number of moles of NaCl:
Molar mass of NaCl = 22.990 g/mol + 35.453 g/mol = 58.443 g/mol
Number of moles of NaCl = 7.000 g / 58.443 g/mol
2. Calculate the number of moles of Pb(NO3)2:
Molar mass of Pb(NO3)2 = 207.200 g/mol + 2 x (14.007 g/mol + 3 x 16.000 g/mol) = 331.209 g/mol
Number of moles of Pb(NO3)2 = 50.400 g / 331.209 g/mol
3. Determine the limiting reagent:
The stoichiometric ratio between NaCl and Pb(NO3)2 is 2:1. Therefore, we need to compare the moles of both substances to determine which one is limiting.
Number of moles of NaCl = 7.000 g / 58.443 g/mol = 0.120 mol
Number of moles of Pb(NO3)2 = 50.400 g / 331.209 g/mol = 0.152 mol
Since the ratio of NaCl to Pb(NO3)2 is 2:1, we can see that NaCl is the limiting reagent because its moles are smaller.
4. Calculate the theoretical yield of PbCl2:
From the stoichiometric ratio 2:1, we know that 2 moles of NaCl will produce 1 mole of PbCl2. Therefore, the moles of PbCl2 produced will be half the moles of NaCl.
Number of moles of PbCl2 = 0.120 mol / 2 = 0.060 mol
5. Calculate the mass of PbCl2 produced:
Molar mass of PbCl2 = 207.200 g/mol + 2 x 35.453 g/mol = 278.106 g/mol
Mass of PbCl2 = 0.060 mol x 278.106 g/mol ≈ 16.686 g
Therefore, the theoretical yield of PbCl2 is approximately 16.686 grams.