Can someone please solve this problem?

Question

Consider the reaction.

2Pb(s)+O2(g)⟶2PbO(s)

An excess of oxygen reacts with 451.4 g of lead, forming 319.3 g of lead(II) oxide. Calculate the percent yield of the reaction.

percent yield:

Answer 1

Good evening Janet,

As we discussed the first time you presented this problem you did not supply the Actual Yield that was needed to complete the problem to determine the % yield. I am glad you found the missing data, so now lets walk through the steps you need to take to find the % yield.

Given: Limiting reactant Pb(s) = 451.4 g

Actual Yield = 319.3 g

Find the % Yield

Equation: 2Pb(s) + O2(g) (in excess) >>>> 2PbO(s)

1. We use moles to do our calculations because if you look at the equation above, it's balanced, meaning the number of moles of Pb as a reactant must equal the number of moles of Pb as a product, and the number of moles of O2 as a reactant must equal the number of moles of O as a product per the law of conversation of mass, mass can neither created or destroyed only changed, Moles is the entity used to do our calculations.

To do stoichiometric calculations as in this problem, we have a given amount of a limiting reactant (Pb) and need to find the number of moles of the product, PbO, which would be your theoretical yield. Theoretical yield means that through calculations you are finding the maximum yield (amount) of the PbO you could obtain. Its theoretical because your determining this by calculations not through experiment. In an actual experiment where you allow this reaction to occur the amount (yield) of PbO you obtain will be the actual yield, the real amount of PbO obtained through experiment and is almost always less than the theoretical yield because you usually lose some of the PbO from spillage, competing reactions, the reaction to form PbO does not go to completion, purification of the PbO causing you to lose some of the product. In this question unlike the last, you suppled the Actual Yield (319.3 g of PbO)

So lets first convert the grams given of Pb(s) to moles:

Molar mass of Pb = 207.2 g

Molar mass of O = 16 g

451.4 g Pb x i mole Pb/207.2 g = (grams cancel out leaving you with moles of Pb) 2.18 mol Pb

2) To determine the moles of product (PbO) we must use the molar ratio of Pb to PbO. Look at the coefficient in front of Pb(s) and the coefficient in front of PbO(s). You should notice that in the balanced equation 2 moles of Pb is required to form 2 moles PbO. a 2:2 ratio or in other words 1:1 ratio, to show you the calculations:

2,18 mol Pb x 2 mol PbO/2 mol Pb the lead cancels leaving you with PbO and just complete the math. 2.18 x 2 mol PbO/2 = 2.18 mol PbO. The Theoretical yield

3) Next we convert the PbO from moles to grams.

molar mass PbO = 223.2 g (the molar mass of Pb + molar mass of O)

2.18 mol PbO x 223.2 g PbO/1 mol PbO = 486.58 g of PbO (the theoretical yield of PbO in grams)

4) last step, calculate the % yield: to do this you take the Actual Yield and divide it by the Theoretical Yield and then multiply by 100 to obtain a percent

319.3 g (Actual Yield)/486.58 g (Theoretical Yield) x 100 = 65.6 % Yield.