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What is the percent yield of NH3 if 23.5 grams NH3 is produced from 13.0 mol H2 and excess N2.

What Is The Percent Yield Of NH3 If 235 Grams NH3 Is Produced From 130 Mol H2 And Excess N2 class=

Sagot :

Given:

actual yield = 23.5 g

moles of H₂ = 13.0 mol

reactants: N₂ and H₂

product: NH₃

Required:

percent yield

Solution:

Step 1: Write the balanced chemical equation.

N₂ + 3H₂ → 2NH₃

Step 2: Calculate the molar mass of NH₃.

molar mass of NH₃ = (14.01 g/mol × 1) + (1.008 g/mol × 3)

molar mass of NH₃ = 17.034 g/mol

Step 3: Calculate the theoretical yield of the reaction by using dimensional analysis (factor-label method).

Since [tex]\text{3 mol} \: \text{H}_{2} \bumpeq \text{2 mol} \: \text{NH}_{3}[/tex], the theoretical yield of the reaction is

[tex]\text{theoretical yield} = \text{13.0 mol} \: \text{H}_{2} × \frac{\text{2 mol} \: \text{NH}_{3}}{\text{3 mol} \: \text{H}_{2}} × \frac{\text{17.034 g} \: \text{NH}_{3}}{\text{1 mol} \: \text{NH}_{3}}[/tex]

[tex]\text{theoretical yield = 147.628 g}[/tex]

Step 4: Calculate the percent yield.

[tex]\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} × 100[/tex]

[tex]\text{percent yield} = \frac{\text{23.5 g}}{\text{147.628 g}} × 100[/tex]

[tex]\boxed{\text{percent yield} = 15.9\%}[/tex]

[tex]\\[/tex]

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