Consider the reaction,
3 H2(g) + N2(g) => 2 NH3(g).
How many grams of NH3 will be made by the reaction of 3 moles of H2(g) with 18 moles of N2(g
Homework Posting Tips
Please show your work. Tutors will not do your homework for you. Please show your work for any question that you are posting.
This is a stoichiometry problem also but in addition it is a limiting reagent problem You know it is a limiting reagent problem because amounts are given for BOTH reactants.
Work this the same way as a stoichiometry problem but take each reactant separately and calculate mols of the product. It is likely you will not get the same value for mols of NH3 with each reactant; the correct value in limiting reagent problems is ALWAYS the smaller value of the product.
To determine the grams of NH3 produced, we first need to know the molar mass of NH3. The molar mass of NH3 is calculated by adding up the atomic masses of the elements in the compound.
The atomic mass of hydrogen (H) is approximately 1 g/mol, and the atomic mass of nitrogen (N) is approximately 14 g/mol. Since NH3 has three hydrogen atoms and one nitrogen atom, the molar mass of NH3 is calculated as follows:
(3 * 1 g/mol) + (1 * 14 g/mol) = 17 g/mol
Now, let's calculate the number of moles of NH3 that can be produced from the given reactants:
From the balanced equation, we can see that the ratio of H2 to NH3 is 3:2. Therefore, for every 3 moles of H2, 2 moles of NH3 will be produced.
Similarly, the ratio of N2 to NH3 is 1:2. For every 1 mole of N2, 2 moles of NH3 will be produced.
Given that we have 3 moles of H2 and 18 moles of N2, we need to find the limiting reactant. This is the reactant that is completely consumed and will determine the amount of product formed.
To determine the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio. The stoichiometric ratio is the ratio of moles required for the reaction.
For H2:
3 moles of H2 * (2 moles of NH3 / 3 moles of H2) = 2 moles of NH3
For N2:
18 moles of N2 * (2 moles of NH3 / 1 mole of N2) = 36 moles of NH3
Since the ratio of H2 to NH3 is lower (2 moles of NH3) compared to the ratio of N2 to NH3 (36 moles of NH3), H2 is the limiting reactant.
Now, let's calculate the mass of NH3 produced using the limiting reactant:
2 moles of NH3 * (17 g/mol) = 34 grams of NH3
Therefore, by the reaction of 3 moles of H2 with 18 moles of N2, 34 grams of NH3 will be produced.