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.    

Cr(oh)3 -> cr(3+) + 3oh-

☆ionisation energy increases across a period from na -> ar☆ionisation energy decreases down a group from li -> fr☆ionic radius decreases across a period from na -> cl☆ionic radius increases down a groupfrom f -> ii hope it clarifies your doubts