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%.

on passing the small ampere of electric current through the tube made of ar and hg of cfl generates uv rays which excites the fluorescent and thus emits the high radiation by consumption of just 5- 12 w in comparison to incandescent light which requires more power such as around 40 - 55 w to lighten.

moreover the life span of fluorescent or cfl is around 8000 h whereas the life span of incandescent is around 1200 h

hence, the building installed with incandescent lights bears more electricity bill.