Using the balanced equation [latex]3H_{2}(g) + N_{2}(g) \longrightarrow 2NH_{3}( g)[/latex] calculate the number of moles of product formed from 5.0 moles of [latex]H_{2} [/latex] and 3.0 moles of [latex]N_{2}.[/latex]

[1] Determine the limiting reactant as in Sample Problem 5.21. • Identify [latex]H_{2}[/latex] as the original quantity and use the mole ratio from the balanced equation as a conversion factor to calculate how much [latex]N_{2}.[/latex] is needed for complete reaction. [latex]\begin{matrix} & & \text{mole–mole conversion factor}& &\\ \\ 5.0 \cancel{mol } H_{2} &\times & \frac{\text{1 mol }N_{2}}{3 \cancel{ mol }{}H_{2}} &= & \text{1.7 mol } N_{2} \text{ needed }\end{matrix} [/latex] • Since the amount of [latex]N_{2}[/latex] present is greater than what is needed, [latex]N_{2}[/latex] is present in excess and [latex]H_{2}[/latex] is the limiting reactant. [2] Convert the number of moles of limiting reactant to the number of moles of product using a mole–mole conversion factor. [latex]\begin{matrix} & & \text{mole–moleconversion factor}& &\\ \\ 5.0 \cancel{mol } H_{2} &\times & \frac{\text{2 mol }NH_{3}}{3 \cancel{ mol }H_{2}} &= & \text{3.3 mol } NH_{3} \text{ formed } \\ &&&& Answer\end{matrix} [/latex]

Question 5.SP.22: Using the balanced equation, 3H2(g) + N2(g)→ 2NH3( g), calcu...

General, organic, & biological chemistry

Using the balanced equation $3H_{2}(g) + N_{2}(g) \longrightarrow 2NH_{3}( g)$ calculate the number of moles of product formed from 5.0 moles of $H_{2}$ and 3.0 moles of $N_{2}.$

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[1] Determine the limiting reactant as in Sample Problem 5.21.
• Identify $H_{2}$ as the original quantity and use the mole ratio from the balanced equation as a conversion factor to calculate how much $N_{2}.$ is needed for complete reaction.

$\begin{matrix} & & \text{mole–mole conversion factor}& &\\ \\ 5.0 \cancel{mol } H_{2} &\times & \frac{\text{1 mol }N_{2}}{3 \cancel{ mol }{}H_{2}} &= & \text{1.7 mol } N_{2} \text{ needed }\end{matrix}$

• Since the amount of $N_{2}$ present is greater than what is needed, $N_{2}$ is present in excess and $H_{2}$ is the limiting reactant.

[2] Convert the number of moles of limiting reactant to the number of moles of product using a mole–mole conversion factor.

$\begin{matrix} & & \text{mole–moleconversion factor}& &\\ \\ 5.0 \cancel{mol } H_{2} &\times & \frac{\text{2 mol }NH_{3}}{3 \cancel{ mol }H_{2}} &= & \text{3.3 mol } NH_{3} \text{ formed } \\ &&&& Answer\end{matrix}$