Calculate the percentage yield for the reaction represented by the equation ch4 + 2o2 ? 2h2o + co2 when 1000 g of ch2 react with excess o2 to produce 2300 g of co2.

83.64%.

Explanation:

∵ The percent yield = (actual yield/theoretical yield)*100.

actual yield of CO₂ = 2300 g.

For the reaction:

CH₄ + 2O₂ → 2H₂O + CO₂,

1.0 mol of CH₄ react with 2 mol of O₂ to produce 2 mol of H₂O and 1.0 mol of CO₂.

no. of moles of CH₄ = mass/molar mass = (1000 g)/(16.0 g/mol) = 62.5 mol.

Using cross-multiplication:

1.0 mol of CH₄ produces → 1.0 mol of CO₂, from stichiometry.

∴ 62.5 mol of CH₄ produces → 62.5 mol of CO₂.

∴ The theoretical yield of CO₂ gas = n*molar mass = (62.5 mol)(44.0 g/mol) = 2750 g.

∵ The percent yield = (actual yield/theoretical yield)*100.

actual yield = 2300 g, theoretical yield = 2750 g.

∴ the percent yield = (2300 g/2750 g)*100 = 83.64%.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.