Verified answer

Lead(II) oxide ....

Heating lead metal in air to a high temperature (~600C) will produce lead(II) oxide, PbO.

2Pb(l) + O2(g) --> 2PbO(s)

451.5g ..................?g

451.5g Pb x (1 mol Pb / 207.2g Pb) x (1 mol PbO / 1 mol Pb) x (223.2g PbO / 1 mol PbO) = 486.4g PbO ............... theoretical yield

Percent yield = 380.9g / 486.4g x 100 = 78.32%

According to the coefficients in the balanced equation, two moles of lead reacts with one mole of oxygen to produce two moles of lead II oxide. The mass of one mole of lead is 207.2 grams.

n = 451.4 ÷ 207.2

The number of moles of lead is approximately 2.2. This is also the number of moles of lead II oxide.

Mass of one mole = 207.2 + 16 = 223.2 grams

To determine the mass of lead II oxide that is produced, multiply the number of moles by the mass of one mole.

Mass = (451.4 ÷ 207.2) * 223.2 = 100,752.48 ÷ 207.2

The mass of lead II oxide is approximately 486 grams.

% yield = 100 * 380.9 ÷ (100,752.48 ÷ 207.2)

This is approximately 78.3% yield. I hope this is helpful for you.

380.9 / 451.4 = 0.853 or 85.3%

that is yield of PbO from Pb

(there are other yields, such as Pb from total reactants)

2Pb + O₂ → 2PbO

2`•207.2 g Pb + 2•16 g O₂ → 2(207.2+16) g PbO

414.4 g Pb + 32 g O₂ → 446.4 g PbO