In the following reaction, 451.4 g of lead reacts with excess oxygen forming 313.6 g of lead(II) oxide. Calculate the percent yield of the reaction.
2Pb(s)+O2(g)= 2PbO(s)
mols Pb = grams/molar mas = ?
Use the coefficients in the balanced equation to convert mols Pb to mols PbO
Now convert mols PbO to g. g = mols x molar mass. This is the theoretical yield (TY)
%yield = (actual yield/TY)*100 = ?
I'm so confused...how do I use the coefficients to balance the equations?
To calculate the percent yield of a reaction, we need to compare the actual yield of the product with the theoretical yield. The actual yield is the amount of product obtained experimentally, while the theoretical yield is the maximum amount of product that can be obtained based on stoichiometry and the amount of limiting reactant.
Let's start by calculating the theoretical yield of lead(II) oxide (PbO) based on the balanced equation.
From the balanced equation, we can see that 2 moles of Pb will react with 1 mole of O2 to produce 2 moles of PbO.
1 mole of PbO has a molar mass of 207.2 g (1 atom of Pb at 207.2 g/mol + 1 atom of O at 16.00 g/mol).
Since the reaction produces 2 moles of PbO for every 2 moles of Pb, we can calculate the theoretical yield of PbO:
Molar mass of PbO = 207.2 g/mol
Theoretical yield of PbO = (451.4 g Pb * 2 mol PbO / 2 mol Pb) * (207.2 g PbO / 1 mol PbO)
Theoretical yield of PbO = 451.4 g PbO
Now we can calculate the percent yield using the actual yield and the theoretical yield:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
In this case, the actual yield is given as 313.6 g of PbO. Plugging in the values, we get:
Percent Yield = (313.6 g PbO / 451.4 g PbO) * 100
Percent Yield = 69.5%
Therefore, the percent yield of the reaction is 69.5%.