When 10.0 g of nh3 reacts, the actual yield of n2 is 8.50 g. what is the percent yield? 4 nh3 (g) + 6 no(g) --> 5 n2 (g) + 6 h2o(l)

The percent yield of N₂ is 41.34%

From the question,

We are to determine the percent yield of N₂

The given balanced chemical equation for the reaction is

4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l)

This means 4 moles of NH₃ reacts with 6 moles of NO to produce 5 moles of N₂ and 6 moles of H₂O

First, we will determine the number of moles of NH₃ that reacted

From the question,

Mass of NH₃ that reacted = 10.0 g

From the formula

Molar mass of NH₃ = 17.031 g/mol

∴ Number of moles of NH₃ that reacted =

Number of moles of NH₃ that reacted = 0.58716 mole

Now, we will determine the theoretical number of moles of N₂ produced

From the balanced chemical equation

Since

4 moles of NH₃ reacts to produce 5 moles of N₂

Then,

0.58716 mole of NH₃ will react to produce moles of N₂

∴ The theoretical number of moles of N₂ that would be produced is 0.73395 mole

Now, we will determine the theoretical yield in grams

Using the formula

Mass = Number of moles × Molar mass

Molar mass of N₂ = 28.0134 g/mol

∴ Mass of N₂ that would be produced = 0.73395 × 28.0134

Mass of N₂ that would be produced = 20.56 g

∴ The theoretical yield of N₂ is 20.56 g

Now, for the percent yield

From the question,

Actual yield of N₂ = 8.50 g

∴ Percent yield of N₂ =

Percent yield of N₂ = 41.34 %

Hence, the percent yield of N₂ is 41.34%

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%Yield = 41.3% (w/w)

Explanation:

4NH₃ + 6NO => 5N₂ + 6H₂O

10g NH₃ = (10/17)mole NH₃ = 0.588 mole NH₃ => 5/4(0.588)mole N₂ = 0.735 mole N₂ x 28g N₂/mole N₂ = 20.58g N₂ (Theoretical Yield)

Given Actual Yield = 8.5g N₂

%Yield = (Actual Yield/Theoretical Yield)100% = (8.5g N₂/20.58g N₂)100% =41.3% Yield  (w/w)

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