N2 + 3 h2 → 2 nh3 . what is the percent yield if 45.5 grams of nh3 is produced upon reaction of 13.0 moles n2 and 29.0 moles h2?

13.82 %.

Explanation:

% yield = [(actual mass of NH₃) / (theoretical mass of NH₃)] x 100.

The actual mass of NH₃ = 45.5 g.

Using cross multiplication:

3.0 moles of H₂ produces → 2.0 moles of NH₃, from the stichiometry,

29.0 moles of H₂ produces → ??? moles of NH₃.

∴ The number of moles of NH₃ produced upon reaction of (13.0 moles N₂ and 29.0 moles H₂) = (2.0)(29.0) / (3.0) = 19.333 mol.

∴ % yield = [(actual mass of NH₃) / (theoretical mass of NH₃)] x 100 = [(45.5 g) / (329.266 g)] x 100 = 13.82 %.

dynamic equilibrium is the answer.

i think that the answer is c