The first step is to determine the theoretical yield of PbCl2 using stoichiometry. The balanced chemical equation for the reaction is:
2 NaCl + Pb(NO3)2 -> 2 NaNO3 + PbCl2
From the equation, we can see that 2 moles of NaCl react with 1 mole of Pb(NO3)2, producing 1 mole of PbCl2. The molar masses of the relevant compounds are:
NaCl: 22.990 + 35.453 = 58.443 g/mol
Pb(NO3)2: 207.200 + 2(14.007 + 3(15.999)) = 331.200 g/mol
PbCl2: 207.200 + 2(35.453) = 278.106 g/mol
Using the masses given, we can calculate the moles of each reactant:
NaCl: 7.000 g / 58.443 g/mol = 0.1197 mol
Pb(NO3)2: 50.400 g / 331.200 g/mol = 0.1520 mol
Since 2 moles of NaCl react with 1 mole of Pb(NO3)2, the limiting reactant is NaCl. Therefore, the moles of PbCl2 produced is:
0.1197 mol NaCl x (1 mol PbCl2/2 mol NaCl) = 0.0598 mol PbCl2
Finally, we can calculate the theoretical yield of PbCl2:
0.0598 mol PbCl2 x 278.106 g/mol = 16.618 g PbCl2
The percent yield is then calculated by dividing the actual yield by the theoretical yield and multiplying by 100%:
percent yield = (12.798 g / 16.618 g) x 100% = 77.343%
Therefore, the answer is 77.343%.
If the theoritical yield of PbCl2 is 16.547 g from the reaction of 7.000 g of NaCl with 50.400 g of Pb(NO3)2. Calculate the % yield of PbCl2 if 12.798 g of PbCl2 was actually obtained? (Atomic mass: Na-22.990 g/mol; 0-15.999 g/mol; Cl-35.453g/mol; Pb-207.200 g/mol; N-14.007).
100%
77.343%
38.343%
60.551%
1 answer
1 answer