Consider the chemical equation. 2h2 + o2 mc022-1.jpg 2h2o what is the percent yield of h2o if 87.0 g of h2o is produced by combining 95.0 g of o2 and 11.0 g of h2? use mc022-2.jpg. 56.5% 59.0% 88.5% 99.7%

88.5 i just took the test

answer: 88.5%

explanation:

1) write the balanced chemical equation:

2h₂ + o₂ → 2h₂o

2) write the mole ratios:

2 mol h₂ : 1 mol o₂ : 2 moles h₂o

3) calculate the number of moles of each reactant:

formula: number of moles = mass in grams / molar massnumber of moles of h₂: 11.0 g / 2.016 g/mol = 5.46 molesnumber of moles of o₂: 95.0 g / 31.998 g/mol = 2.97 moles

4) determine the limiting reactant:

theoretical ratio: 1 mol o₂ / 2 mol h₂ = 0.5 o₂ / h₂actual ratio: 2.97 mol o₂ / 5.46 mol h₂ ≈ 0.54 o₂ / h₂hence, there is more o₂ moles than the needed to react with the h₂ moles given, so o₂ is in excess and h₂ is the limiting reactant.

5) calculate the theoretical yield using the limiting reactant:

proportion: 2 moles h₂ / 2 moles h₂o = 5.46 moles h₂ / xsolve for x: x = 5.46 moles of h₂oconvert moles to mass of h₂o: 5.46 mol × 18.015 g / mol = 98.4 g

6) calculate the percent yield:

percent yiled = (actual yield / theoretical yield) × 100percent yield = (87.0 g / 98.4 g) × 100 = 88.4% ← answer

note: the 0.1% difference of 88.4% with 88.5% is not relevant since the difference is in the last significant figure and it can be attributed to differences in the precision in the molar masses used.    

Heating it up with cause a quicker reaction time

Answerlimiting reactant;