Assume that you have 1.41mol of H2 and 3.50mol of N2. How many grams of ammonia (NH3) can you make, and how many grams of which reactant will be left over?
3H2+N2→2NH3
A limiting reagent problem with an extra twist at the end.
Write and balance the equation
N2 + 3H2 ==> 2NH3
Using the coefficients in the balanced equation, convert mols N2 and mols H2 to mols NH3.
N2 first:
1.41 mol N2 x (2 mol NH3/1 mol N2) = 2.82 mols NH3 produced (if we had 1.41 mol N2 and all of the H2 we needed.)
H2 next:
3.50 mol H2 x (2 mol NH3/3 mol H2) = 2.33 mol NH3 produced (if we had 3.50 mol H2 and all of the N2 we needed.)
You obtained two values and you know that can't be right; one of them must be wrong. In limiting reagent problems the smaller values is ALWAYS the correct one and the reagent producing that value is the limiting reagent. Therefore, you will have 2.33 mol NH3 produced and H2 is the limiting reagent.
To convert mols NH3 to grams it is g = mols x molar mass..
To find the mols of the "other" reagent(the non-limiting reagent or the excess reagent), use the coefficients again.
3.5 mol H2 x (1 mol N2/3 mol H2) = 1.17 mol N2 used.
1.41-1.17 = 0.24 mols N2 left un-reacted.
That will be 0.24 x 28 = about 6.8 grams N2 remaining.
To solve this problem, we need to first balance the chemical equation:
3H2 + N2 → 2NH3
Now, let's find the limiting reactant:
1. Calculate the number of moles of NH3 that can be produced from each reactant:
- From H2: 1.41 mol H2 * (2 mol NH3 / 3 mol H2) = 0.94 mol NH3
- From N2: 3.50 mol N2 * (2 mol NH3 / 1 mol N2) = 7.00 mol NH3
2. Compare the mole ratios of NH3 produced from each reactant:
The mole ratio of NH3 to H2 is 2:3, and the mole ratio of NH3 to N2 is 2:1. Since the mole ratio of NH3 to H2 is less than the mole ratio of NH3 to N2, H2 is the limiting reactant.
3. Calculate the amount of NH3 produced from the limiting reactant:
0.94 mol NH3
4. Now, let's calculate the reactant left over. To do that, we need to determine the number of moles of H2 remaining:
- Moles of H2 used = 1.41 mol H2
- Moles of NH3 produced from H2 = 0.94 mol NH3
- Moles of H2 remaining = Moles of H2 used - Moles of NH3 produced from H2 = 1.41 mol H2 - 0.94 mol NH3 = 0.47 mol H2
To convert the remaining moles of H2 to grams, we need to use its molar mass:
5. Calculate the mass of H2 left over:
- Molar mass of H2 = 2 g/mol
- Mass of H2 left over = Moles of H2 remaining * Molar mass of H2 = 0.47 mol H2 * 2 g/mol = 0.94 g H2
Therefore, you can make 0.94 g of ammonia (NH3), and there will be 0.94 g of hydrogen (H2) left over.
To determine the grams of ammonia (NH3) that can be produced and the grams of the reactant that will be left over, we need to follow these steps:
1. Convert the number of moles of H2 and N2 to moles of NH3 using the stoichiometric coefficients from the balanced equation: 3H2 + N2 → 2NH3
- Given 1.41 mol of H2, we can calculate the moles of NH3 produced:
* 1.41 mol H2 * (2 mol NH3 / 3 mol H2) = 0.940 mol NH3
- Given 3.50 mol of N2, we can calculate the moles of NH3 produced:
* 3.50 mol N2 * (2 mol NH3 / 1 mol N2) = 7.00 mol NH3
2. Determine the limiting reactant (the reactant that is completely consumed) by comparing the moles of NH3 produced from each reactant. The reactant that produces fewer moles of NH3 is the limiting reactant.
- In this case, the H2 produces 0.940 mol of NH3, while the N2 produces 7.00 mol of NH3. Therefore, H2 is the limiting reactant.
3. Calculate the theoretical yield of NH3 based on the limiting reactant. The theoretical yield is the maximum amount of product that can be obtained from the limiting reactant.
- Using the limiting reactant (H2), calculate the grams of NH3 produced:
* 0.940 mol NH3 * (17.031 g NH3 / 1 mol NH3) = 15.99 g NH3
4. Determine the excess reactant (the reactant that remains after the limiting reactant is completely consumed) and calculate the amount left over.
- To find the moles of excess reactant, subtract the moles of NH3 produced from the total moles of the reactant:
* Moles of excess H2 = 1.41 mol H2 - 0.940 mol NH3 * (3 mol H2 / 2 mol NH3) = 1.12 mol H2
- Convert the moles of excess H2 to grams:
* 1.12 mol H2 * (2.016 g H2 / 1 mol H2) = 2.26 g H2
Therefore, you can produce 15.99 grams of NH3, and 2.26 grams of H2 will be left over.