Question

ammonia (nh₃) can be produced by the reaction of hydrogen gas with nitrogen gas:
3h₂ (g) + n₂ (g) ⟶ 2nh₃ (g)
a chemist reacts 2.00 mol h₂ with excess n₂. the reaction yields 0.80 mol nh₃.
what is the percent yield of the reaction?

○ 40%
○ 60%
○ 80%
○ 25%

question 17
1 pts
what would cause the percent yield of a reaction to be less than 100%

○ all of the limiting reagent gets used up.
○ a balanced chemical equation is used.
○ the product participates in a side reaction.
○ pure reactants are used.

Explanation:

Step1: Calculate the theoretical yield of \(NH_3\)

From the balanced chemical equation \(3H_2(g)+N_2(g)\to2NH_3(g)\), the mole ratio of \(H_2\) to \(NH_3\) is \(3:2\).
Given \(n(H_2) = 2.00\space mol\), using the mole - ratio \(\frac{n(NH_3)}{n(H_2)}=\frac{2}{3}\), so \(n_{theoretical}(NH_3)=\frac{2}{3}\times n(H_2)\)
\(n_{theoretical}(NH_3)=\frac{2}{3}\times2.00\space mol=\frac{4}{3}\space mol\approx1.33\space mol\)

Step2: Calculate the percent yield

The formula for percent yield is \(\text{Percent Yield}=\frac{n_{actual}(NH_3)}{n_{theoretical}(NH_3)}\times100\%\)
Given \(n_{actual}(NH_3) = 0.80\space mol\) and \(n_{theoretical}(NH_3)\approx1.33\space mol\)
\(\text{Percent Yield}=\frac{0.80\space mol}{\frac{4}{3}\space mol}\times 100\%=\frac{0.80\times3}{4}\times100\% = 60\%\)

For the second question:

• If all of the limiting reagent gets used up (\(n_{theoretical}\) is calculated based on complete consumption of limiting reagent), and a balanced chemical equation is used (to calculate \(n_{theoretical}\) correctly) and pure reactants are used (no impurity to affect the reaction), these are conditions for a high - yield or \(100\%\) yield (in theory).
• If the product participates in a side reaction, some of the product is consumed in another reaction, so the actual yield (\(n_{actual}\)) is less than the theoretical yield (\(n_{theoretical}\)), which causes the percent yield (\(\text{Percent Yield}=\frac{n_{actual}}{n_{theoretical}}\times100\%\)) to be less than \(100\%\)

Answer:

First question: \(60\%\) (the second option in the first multiple - choice question)
Second question: The product participates in a side reaction (the third option in the second multiple - choice question)