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2 C6H5NO2 + 4 C6H14O4 ® (c6h5n)2 + 4 C6H12O4 + 4 H2O what percent yield of this reaction given that the in one reaction, 0.10 l of nitrobenzene (d = 1.20 g/ml) and 0.30 L of triethylene glycol (d = 1.12 g/ml) yields 55 g azobenzene.
Accepted Answer
Here's how to calculate the percent yield:
1. Calculate the mass of nitrobenzene:
Volume = 0.10 L = 100 mL
Mass = Volume x Density = 100 mL x 1.20 g/mL = 120 g
2. Calculate the mass of triethylene glycol:
Volume = 0.30 L = 300 mL
Mass = Volume x Density = 300 mL x 1.12 g/mL = 336 g
3. Determine the limiting reactant:
Calculate the moles of each reactant:
Moles of nitrobenzene = 120 g / (123.11 g/mol) = 0.975 mol
Moles of triethylene glycol = 336 g / (150.17 g/mol) = 2.24 mol
The stoichiometry of the reaction shows that 2 moles of nitrobenzene react with 4 moles of triethylene glycol. Therefore, nitrobenzene is the limiting reactant (0.975 mol nitrobenzene would require 1.95 mol triethylene glycol).
4. Calculate the theoretical yield of azobenzene:
From the balanced equation, 2 moles of nitrobenzene produce 1 mole of azobenzene.
Theoretical yield = (0.975 mol nitrobenzene)
(1 mol azobenzene / 2 mol nitrobenzene)
(182.22 g/mol azobenzene) = 88.9 g azobenzene
5. Calculate the percent yield:
Percent Yield = (Actual Yield / Theoretical Yield) x 100%
Percent Yield = (55 g / 88.9 g) x 100% = 61.9%
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