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Having problems staying logged in or are you experiencing issues? Please visit our tre A student is using these conversion factors to convert from 5.0 moles of the given reactant to the moles of product. Idenfify any conversion factors that would NOT work. Tap to select or deselect an answer. Select all that would NOT work. \[ \begin{array}{l} \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} \\ 5.0 \mathrm{~mol} \mathrm{H}_{2} \times \frac{2 \mathrm{~mol} \mathrm{NH}_{3}}{3 \mathrm{~mol} \mathrm{H}_{2}} \end{array} \] \( 3 \mathrm{Ti}+2 \mathrm{~N}_{2} \rightarrow \mathrm{Ti}_{3} \mathrm{~N}_{4} \) \[ 3 \mathrm{~mol} \mathrm{Ti} \times \frac{1 \mathrm{~mol} \mathrm{Ti}_{3} \mathrm{~N}_{4}}{5.0 \mathrm{~mol} \mathrm{Ti}^{\prime}} \] \[ 2 \mathrm{Al}+3 \mathrm{Br}_{2} \rightarrow 2 \mathrm{AlBr}_{3} \] \( 5.0 \mathrm{~mol} \mathrm{Br}_{2} \times \frac{2 \mathrm{~mol} \mathrm{Br}_{2}}{3 \mathrm{~mol} \mathrm{AlBr}_{3 \mathrm{r}}} \)

Ask by Johnson Zimmerman. in the United States
Feb 18,2025

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The conversion factors that would NOT work are: - \( 3 \mathrm{~mol} \mathrm{Ti} \times \frac{1 \mathrm{~mol} \mathrm{Ti}_{3} \mathrm{~N}_{4}}{5.0 \mathrm{~mol} \mathrm{Ti}} \) - \( 5.0 \mathrm{~mol} \mathrm{Br}_{2} \times \frac{2 \mathrm{~mol} \mathrm{Br}_{2}}{3 \mathrm{~mol} \mathrm{AlBr}_{3}} \)

Solución

We’re given three different reactions with corresponding conversion‐factors that a student has written. The goal is to convert 5.0 moles of a given reactant into moles of product. However, some of the conversion factors written by the student are set up improperly. Let’s examine each one. ────────────────────────────── 1) For the reaction   N₂ + 3 H₂ → 2 NH₃ the student writes:   5.0 mol H₂ × (2 mol NH₃ / 3 mol H₂) Analysis: • In the balanced equation, 3 mol H₂ produce 2 mol NH₃. • Using H₂ as the starting reactant, the proper conversion factor is (2 mol NH₃ / 3 mol H₂). • Units cancel correctly: mol H₂ cancels and you obtain mol NH₃. • This conversion factor is set up correctly. ────────────────────────────── 2) For the reaction   3 Ti + 2 N₂ → Ti₃N₄ the student writes:   3 mol Ti × (1 mol Ti₃N₄ / 5.0 mol Ti) Analysis: • The balanced equation shows that 3 mol Ti produce 1 mol Ti₃N₄. • The correct conversion factor should be (1 mol Ti₃N₄ / 3 mol Ti). • Instead, the student’s conversion factor places 5.0 mol Ti in the denominator. • (Also note that “5.0” appears to be out of place because we’re converting moles that were given, and the stoichiometric coefficient for Ti is 3, not 5.0.) • Thus, this conversion factor is written incorrectly and would not work. ────────────────────────────── 3) For the reaction   2 Al + 3 Br₂ → 2 AlBr₃ the student writes:   5.0 mol Br₂ × (2 mol Br₂ / 3 mol AlBr₃) Analysis: • In the balanced equation, 3 mol Br₂ yield 2 mol AlBr₃. • When converting from moles of Br₂ (the given reactant) to moles of product, the proper conversion factor is:   (2 mol AlBr₃ / 3 mol Br₂) • However, the student’s conversion factor is (2 mol Br₂ / 3 mol AlBr₃) – note that the moles for Br₂ and AlBr₃ are swapped. • With the student’s factor, you would end up with units of Br₂ rather than AlBr₃. • Thus, this conversion factor is also set up the wrong way; it would not work. ────────────────────────────── Summary of the Incorrect Conversion Factors: • The conversion factor for the titanium reaction (3 Ti + 2 N₂ → Ti₃N₄) is incorrect because it uses “5.0 mol Ti” instead of “3 mol Ti” in the denominator. • The conversion factor for the aluminum bromide reaction (2 Al + 3 Br₂ → 2 AlBr₃) is incorrect because the ratio is inverted (moles of Br₂ appear in the numerator rather than moles of AlBr₃). ────────────────────────────── Answer: The conversion factors that would NOT work are: • 3 mol Ti × (1 mol Ti₃N₄ / 5.0 mol Ti) • 5.0 mol Br₂ × (2 mol Br₂ / 3 mol AlBr₃) These are the ones you should deselect.

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Beyond the Answer

In the equation with the nitrogen and hydrogen, using the conversion factor \(\frac{2 \mathrm{~mol} \mathrm{NH}_{3}}{3 \mathrm{~mol} \mathrm{H}_{2}}\) works perfectly since it maintains the stoichiometric ratios from the balanced equation. However, the conversion factor for titanium \(\frac{1 \mathrm{~mol} \mathrm{Ti}_{3} \mathrm{~N}_{4}}{5.0 \mathrm{~mol} \mathrm{Ti}^{\prime}}\) does not align because it misrepresents the molar ratio needed for the conversion and does not come from the balanced equation. For the aluminum and bromine reaction, the conversion factor \(\frac{2 \mathrm{~mol} \mathrm{Br}_{2}}{3 \mathrm{~mol} \mathrm{AlBr}_{3}}\) is incorrect since it suggests that two moles of bromine create three moles of aluminum bromide, which is the reverse of the actual stoichiometry. Always ensure the numerator and denominator accurately reflect the expected products and reactants for effective conversions!

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