To determine the volume of hydrogen gas ([tex]\(H_2\)[/tex]) needed to produce 446 liters of ammonia ([tex]\(NH_3\)[/tex]) at STP according to the chemical equation:
[tex]\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \][/tex]
follow these steps:
1. Understand the balanced chemical equation: This tells us the stoichiometric relationship between the reactants and products. Specifically, 1 mole (or volume) of nitrogen gas ([tex]\(N_2\)[/tex]) reacts with 3 moles (or volumes) of hydrogen gas ([tex]\(H_2\)[/tex]) to produce 2 moles (or volumes) of ammonia ([tex]\(NH_3\)[/tex]).
2. Identify the ratio: From the balanced equation, 3 volumes of [tex]\(H_2\)[/tex] produce 2 volumes of [tex]\(NH_3\)[/tex]. This can be expressed as:
[tex]\[ \frac{3 \text{ volumes } H_2}{2 \text{ volumes } NH_3} \][/tex]
3. Given data: You are given that 446 liters of [tex]\(NH_3\)[/tex] are produced.
4. Set up the proportion: To find the volume of [tex]\(H_2\)[/tex] needed, use the stoichiometric ratio:
[tex]\[ \text{Volume of } H_2 = \left( \frac{3 \text{ volumes } H_2}{2 \text{ volumes } NH_3} \right) \times \text{volume of } NH_3 \][/tex]
5. Plug in the known value: Replace the volume of [tex]\(NH_3\)[/tex] with 446 liters:
[tex]\[ \text{Volume of } H_2 = \left( \frac{3}{2} \right) \times 446 \text{ liters} \][/tex]
6. Calculate the result:
[tex]\[ \text{Volume of } H_2 = \frac{3}{2} \times 446 \][/tex]
[tex]\[ \text{Volume of } H_2 = 1.5 \times 446 \][/tex]
[tex]\[ \text{Volume of } H_2 = 669 \text{ liters} \][/tex]
Therefore, the volume of hydrogen gas needed to produce 446 liters of ammonia at STP is 669 liters.
