The balanced chemical equation for the reaction is:
$$N_2 (g) + 3H_2 (g) \rightarrow 2NH_3 (g)$$
From the equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
Given that we have 28 g of nitrogen, let's convert it to moles:
$$Moles \ of \ N_2 = 28 \ g / 28 \ g/mol = 1 \ mol$$
Now, let's assume we have an excess of hydrogen gas. To determine the maximum amount of ammonia that can be produced, we will use the mole ratio from the balanced equation.
$$Moles \ of \ NH_3 \ produced = 1 \ mol \ N_2 \times \frac{2 \ mol \ NH_3}{1 \ mol \ N_2} = 2 \ mol \ NH_3$$
Finally, we convert the moles of ammonia back to grams:
$$Grams \ of \ NH_3 = 2 \ mol \ NH_3 \times 17 \ g/mol = 34 \ g$$
Therefore, if 28 g of nitrogen are available, the maximum amount of ammonia that can be produced is 34 grams.
