$$N_2 + 3H_2 \rightarrow 2NH_3$$
To determine the mass at the end of the reaction, we need to calculate the limiting reactant. This is the reactant that is completely consumed in the reaction, limiting the amount of product that can be formed.
To calculate the limiting reactant, we can compare the actual mole ratios of the reactants to the stoichiometric mole ratios from the balanced chemical equation.
First, we calculate the moles of each reactant:
$$Moles \ of \ N_2 = 45 g / 28 g/mol = 1.61 mol$$
$$Moles \ of \ H_2 = 30 g / 2 g/mol = 15 mol$$
Next, we calculate the mole ratio of the reactants:
$$Mole \ ratio \ of \ N_2 \ to \ H_2 = 1.61 mol / 15 mol = 0.107$$
The stoichiometric mole ratio of N2 to H2 from the balanced chemical equation is 1:3, which is equivalent to 0.333.
Comparing the actual mole ratio to the stoichiometric mole ratio, we can see that N2 is the limiting reactant because its actual mole ratio is less than the stoichiometric mole ratio. This means that all of the N2 will be consumed in the reaction, and the amount of NH3 produced will be limited by the amount of N2 available.
To calculate the mass of NH3 produced, we use the stoichiometry of the balanced chemical equation. For every 1 mole of N2 that reacts, 2 moles of NH3 are produced. The molar mass of NH3 is 17 g/mol.
$$Moles \ of \ NH_3 \ produced = 1.61 mol \ N_2 \times 2 mol \ NH_3 / 1 mol \ N_2 = 3.22 mol \ NH_3$$
$$Mass \ of \ NH_3 \ produced = 3.22 mol \ NH_3 \times 17 g/mol = 54.54 g$$
Therefore, the mass at the end of the reaction will be 54.54 g of NH3.
