What is the limiting reagent when 150.0 g of nitrogen reacts with 32.1 g of hydrogen?
N2(g) + 3H(g) > 2NH3(g)
First, fix your equation
how many moles of each reagent do you have?
If moles(H2) is less than 3 times the moles(N2) then H is the limiting reagent.
To determine the limiting reagent, we need to compare the number of moles of nitrogen (N2) and hydrogen (H2) in the given amounts.
1. Convert the masses of each element to moles:
- Moles of nitrogen (N2): 150.0 g / molar mass of N2 = x moles
- Moles of hydrogen (H2): 32.1 g / molar mass of H2 = y moles
2. Set up a mole ratio using the balanced chemical equation:
N2(g) + 3H2(g) → 2NH3(g)
According to the balanced equation, we need 1 mole of N2 for every 3 moles of H2.
3. Compare the moles of N2 and H2 using the mole ratio from the balanced equation:
- N2 moles / N2 coefficient = x moles / 1
- H2 moles / H2 coefficient = y moles / 3
4. Determine which reactant has the lowest mole ratio. The reactant with the lower mole ratio is the limiting reagent.
In this case, compare x/1 and y/3 to find the lowest value.
- If x/1 < y/3, then nitrogen (N2) is the limiting reagent.
- If x/1 > y/3, then hydrogen (H2) is the limiting reagent.
Perform the calculation to determine the limiting reagent.
To determine the limiting reagent in a chemical reaction, you need to compare the amount of each reactant to the stoichiometric ratio given in the balanced equation. The reactant that yields the smaller amount of product is the limiting reagent because it will be completely consumed, limiting the amount of product formed.
Let's start by calculating the number of moles for each reactant using their respective masses and molar masses:
Given:
Mass of nitrogen (N2) = 150.0 g
Molar mass of nitrogen (N2) = 28.02 g/mol
Number of moles of nitrogen (N2):
150.0 g / 28.02 g/mol = 5.35 mol
Given:
Mass of hydrogen (H2) = 32.1 g
Molar mass of hydrogen (H2) = 2.02 g/mol
Number of moles of hydrogen (H2):
32.1 g / 2.02 g/mol = 15.89 mol
Now that we have the number of moles for each reactant, we can determine the stoichiometric ratio in which they react. From the balanced equation, we see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Stoichiometric ratio of nitrogen (N2) to hydrogen (H2):
N2:H2 = 1:3
If the number of moles of hydrogen is larger than three times the number of moles of nitrogen, then nitrogen would be the limiting reagent. If the number of moles of hydrogen is smaller, then hydrogen would be the limiting reagent.
Given:
Number of moles of nitrogen (N2) = 5.35 mol
Number of moles of hydrogen (H2) = 15.89 mol
Since 15.89 mol > 3 * 5.35 mol, we can conclude that nitrogen (N2) is the excess reagent and hydrogen (H2) is the limiting reagent in this reaction.
Therefore, hydrogen (H2) is the limiting reagent when 150.0 g of nitrogen reacts with 32.1 g of hydrogen.