What is the limiting reagent when 150.0 g of nitrogen react with 32.1 g of hydrogen?
N2(g) + 3H(g) > 2NH3(g)
Thank you
Here is a worked example of a limiting reagent.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
This didn’t even solve the right problem. It solved a totally different problem...
To determine the limiting reagent, we have to calculate the amount of product that can be formed from each reactant and compare them.
1. First, let's convert the mass of each reactant to moles using their respective molar masses:
- Nitrogen (N2):
150.0 g N2 × (1 mol N2 / 28.02 g N2) = 5.357 mol N2
- Hydrogen (H2):
32.1 g H2 × (1 mol H2 / 2.016 g H2) = 15.926 mol H2
2. Next, let's determine the number of moles of product formed from each reactant, using the balanced equation:
From the balanced equation, we know that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Therefore:
- For Nitrogen:
5.357 mol N2 × (2 mol NH3 / 1 mol N2) = 10.714 mol NH3
- For Hydrogen:
15.926 mol H2 × (2 mol NH3 / 3 mol H2) = 10.617 mol NH3
3. Finally, we compare the number of moles of NH3 produced from each reactant. The reactant that produces the lesser amount of NH3 is the limiting reagent:
In this case, hydrogen (H2) produces 10.617 mol of NH3, while nitrogen (N2) produces 10.714 mol of NH3. Therefore, hydrogen is the limiting reagent.
Thus, hydrogen is the limiting reagent when 150.0 g of nitrogen reacts with 32.1 g of hydrogen in this reaction.
To determine the limiting reagent in a chemical reaction, you need to compare the moles of the available reactants and see which one will run out first. The reactant that gets completely consumed is the limiting reagent.
To find the limiting reagent, follow these steps:
Step 1: Convert the mass of each reactant to moles.
Given:
Mass of nitrogen (N2) = 150.0 g
Mass of hydrogen (H2) = 32.1 g
To convert mass to moles, divide each mass by the molar mass.
The molar mass of nitrogen (N2) is 28.02 g/mol, and the molar mass of hydrogen (H2) is 2.02 g/mol.
Number of moles of nitrogen (N2):
150.0 g / 28.02 g/mol = 5.35 mol
Number of moles of hydrogen (H2):
32.1 g / 2.02 g/mol = 15.89 mol
Step 2: Determine the stoichiometric ratio of the reactants.
According to the balanced equation:
N2(g) + 3H2(g) → 2NH3(g)
The ratio between nitrogen (N2) and hydrogen (H2) is 1:3.
Step 3: Calculate the moles of nitrogen required to react with the available amount of hydrogen.
Since the stoichiometric ratio between nitrogen (N2) and hydrogen (H2) is 1:3, for every 3 moles of hydrogen, 1 mole of nitrogen is required.
Therefore, the moles of nitrogen required to react with the available hydrogen is:
15.89 mol / 3 = 5.30 mol
Step 4: Compare the calculated moles of nitrogen required to the moles of nitrogen available.
The calculated moles of nitrogen required is 5.30 mol, and the moles of nitrogen available is 5.35 mol.
Since the calculated moles of nitrogen required is slightly less than the moles of nitrogen available, the limiting reagent is nitrogen (N2).
Therefore, nitrogen (N2) is the limiting reagent in this reaction.