To determine the volume of hydrogen (H[tex]\(_2\)[/tex]) needed to generate 446 liters of ammonia (NH[tex]\(_3\)[/tex]) at Standard Temperature and Pressure (STP), we can use the stoichiometry of the balanced chemical reaction:
[tex]\[
N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g)
\][/tex]
Here's a step-by-step solution:
1. Identify the stoichiometric relationship:
According to the balanced equation, 1 mole of [tex]\(\text{N}_2\)[/tex] reacts with 3 moles of [tex]\(\text{H}_2\)[/tex] to produce 2 moles of [tex]\(\text{NH}_3\)[/tex].
2. Establish the volume ratio:
At STP, gases react in volumes proportional to their stoichiometric coefficients. Therefore, the volume ratio between [tex]\(\text{H}_2\)[/tex] and [tex]\(\text{NH}_3\)[/tex] can be derived from their coefficients in the balanced equation:
[tex]\[
\frac{3 \text{ volumes of } \text{H}_2}{2 \text{ volumes of } \text{NH}_3}
\][/tex]
3. Apply the volume ratio:
Now, use the given volume of [tex]\(\text{NH}_3\)[/tex] to find the required volume of [tex]\(\text{H}_2\)[/tex]. Let's denote the volume of [tex]\(\text{H}_2\)[/tex] needed as [tex]\(V_{\text{H}_2}\)[/tex]:
[tex]\[
\frac{V_{\text{H}_2}}{446\ \text{L of NH}_3} = \frac{3}{2}
\][/tex]
4. Solve for [tex]\(V_{\text{H}_2}\)[/tex]:
Multiply both sides of the equation by 446 L to solve for [tex]\(V_{\text{H}_2}\)[/tex]:
[tex]\[
V_{\text{H}_2} = 446\ \text{L} \times \frac{3}{2}
\][/tex]
Simplifying this:
[tex]\[
V_{\text{H}_2} = 669\ \text{L}
\][/tex]
Therefore, the volume of hydrogen needed to generate 446 liters of ammonia at STP is 669 liters.
