Respuesta :
Answer:
%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)
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
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
Molar mass of NH₃ = 17.031 g/mol
∴ Number of moles of NH₃ that reacted = [tex]\frac{10.0}{17.031 }[/tex]
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 [tex]\frac{0.58716 \times 5}{4}[/tex] moles of N₂
[tex]\frac{0.58716 \times 5}{4} = 0.73395[/tex]
∴ 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
[tex]Percent\ yield = \frac{Actual\ yield}{Theoretical\ yield}\times 100\%[/tex]
From the question,
Actual yield of N₂ = 8.50 g
∴ Percent yield of N₂ = [tex]\frac{8.50}{20.56}\times 100\%[/tex]
Percent yield of N₂ = 41.34 %
Hence, the percent yield of N₂ is 41.34%
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