Questions: If 28 moles of NO combine with 45 moles of H₂, how many moles of NH₃ can form? What is the limiting reagent? 2 NO(g) + 5 H₂(g) → 2 NH₃(g) + 2 H₂O(g) Step 1: Show how you would determine the number of moles of NH₃ that could form from each starting material. 28 moles NO × () = ? moles NH₃ 45 moles H₂ × () = ? moles NH₃ Answer Bank 2 moles NO 5 moles H₂ 2 moles NH₃ 2 moles H₂O
Transcript text: If 28 moles of NO combine with 45 moles of $\mathrm{H}_{2}$, how many moles of $\mathrm{NH}_{3}$ can form? What is the limiting reagent?
\[
2 \mathrm{NO}(\mathrm{~g})+5 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g})
\]
Step 1: Show how you would determine the number of moles of $\mathrm{NH}_{3}$ that could form from each starting material.
\[
\begin{array}{l}
28 \text { moles } \mathrm{NO} \times \frac{\square}{\square}=\text { ? moles } \mathrm{NH}_{3} \\
45 \text { moles } \mathrm{H}_{2} \times \frac{\square}{\square}=\text { ? moles } \mathrm{NH}_{3}
\end{array}
\]
Answer Bank
2 moles NO
5 moles $\mathrm{H}_{2}$
2 moles $\mathrm{NH}_{3}$
2 moles $\mathrm{H}_{2} \mathrm{O}$
Solution
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