N2(g) + 3 H2(g) → 2 NH3(g) [balanced]
If 5.42 g of nitrogen gas are reacted with 5.42 g of hydrogen gas, which of the reactants is the limiting reactant? Use
the molar mass data below if necessary.
Molar mass of N2 = 28.02 g/mol
Molar mass of H2 = 2.02 g/mol
Molar mass of NH3 = 17.04 g/mo
5.42 grams H2=2.71 moles
YOu need 1/3 of that of N2
moles N2=1/3*2.71=.89
grams N2=.89*28= way more than the 5.42 grams you have, so N2 is the limiting reageant.
Convert gram values given to moles. (moles = mass / f.wt), then divide the mole values by respective coefficients... the smaller number is the limiting reagent. Moles N2 = 5.42/28.02 = 0.193; Moles H2 = 5.42/2.02 = 2.68 => compare N2(0.193/1) and H2(2.68/3) = 0.894... N2 is the limiting reagent & H2 will be in excess.
To determine the limiting reactant, we need to calculate the amount of each reactant in moles and compare them to the stoichiometry of the balanced equation.
Given:
Mass of N2 = 5.42 g
Mass of H2 = 5.42 g
First, let's convert the masses of N2 and H2 to moles using the molar masses given:
Moles of N2 = Mass of N2 / Molar mass of N2
Moles of N2 = 5.42 g / 28.02 g/mol
Moles of H2 = Mass of H2 / Molar mass of H2
Moles of H2 = 5.42 g / 2.02 g/mol
Calculating these values:
Moles of N2 = 0.1936 mol
Moles of H2 = 2.68 mol
Now, let's look at the stoichiometry of the balanced equation: N2(g) + 3 H2(g) → 2 NH3(g).
According to the balanced equation, the ratio of moles of N2 to moles of H2 is 1:3. This means that for every mole of N2, three moles of H2 are required.
To determine the limiting reactant, we need to see which reactant has a lesser number of moles compared to the stoichiometric ratio.
For N2:
Moles of N2 = 0.1936 mol
For H2:
Moles of H2 = 2.68 mol
Comparing the values to the stoichiometric ratio:
N2: 0.1936 mol
H2: 2.68 mol * (1/3) = 0.8933 mol
From the comparison, we can see that the number of moles of N2 (0.1936 mol) is lesser than the number of moles of H2 (0.8933 mol).
Therefore, the limiting reactant is N2, as it is completely consumed before H2 in the given reaction.
To determine the limiting reactant, we need to compare the number of moles of each reactant with the stoichiometric ratio in the balanced equation.
First, we need to calculate the number of moles of each reactant:
Moles of N2 = Mass of N2 / Molar mass of N2
= 5.42 g / 28.02 g/mol
= 0.1937 mol
Moles of H2 = Mass of H2 / Molar mass of H2
= 5.42 g / 2.02 g/mol
= 2.6832 mol
Next, we need to calculate the stoichiometric ratio between N2 and H2 based on the balanced equation. From the equation, we can see that the ratio is 1:3. This means that for every 1 mole of N2, we need 3 moles of H2.
Now, let's compare the moles of N2 and H2 to the stoichiometric ratio:
Moles of N2 / Stoichiometric ratio = 0.1937 mol / 1 = 0.1937 mol
Moles of H2 / Stoichiometric ratio = 2.6832 mol / 3 = 0.8944 mol
As we can see, the moles of N2 divided by the stoichiometric ratio is lower than the moles of H2 divided by the stoichiometric ratio. This indicates that N2 is the limiting reactant because there is not enough N2 to completely react with the available H2.
Therefore, in this reaction, N2 is the limiting reactant.