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:

The % yield is calculated from the relation:

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

The actual mass of NH₃ = 45.5 g.

Now, we should calculate the theoretical mass of NH₃ produced upon reaction of 13.0 moles N₂ and 29.0 moles H₂.From the balanced equation (N₂ + 3H₂ → 2NH₃), it is clear that every 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.The limiting reactant is H₂ (29.0 moles) that requires 9.667 moles of N₂ (excess) to be completely reacted.

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.

Now, we can calculate the theoretical mass of NH₃ produced = n x molar mass = (19.333 mole)(17.031 g/mol)  = 329.266 g.

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

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Sorry i dont understand the question can you rephase it?