5.22kg of H2 reacts with 31.5kg of N2, which one is the limiting reactant?
I assume we are talking about:
3 H2 + N2 --> 2 NH3
H2 is 2*1 = 2 g/mol
N2 is 2*14 = 28 g/mol
forget g vs kg because we are only interested in ratios here
3 mol H2 is 6 g
so we need 6 g of H2 for every 28 g of N2
31.5 g of N2 needs 31.5 * 6/28 = 6.75 g H2 needed
BUT we only have 5.22 g H2 so the H2 limits the reaction.
To determine the limiting reactant, we need to compare the given masses of each reactant (H2 and N2) with their respective molar masses.
First, let's calculate the number of moles for each reactant using the equation: moles = mass / molar mass.
The molar mass of H2 is 2 g/mol (2 hydrogen atoms x 1 g/mol for each atom).
The molar mass of N2 is 28 g/mol (2 nitrogen atoms x 14 g/mol for each atom).
For H2:
moles of H2 = mass of H2 / molar mass of H2
moles of H2 = 5.22 kg / 2 g/mol
moles of H2 = 5220 g / 2 g/mol
moles of H2 = 2610 mol
For N2:
moles of N2 = mass of N2 / molar mass of N2
moles of N2 = 31.5 kg / 28 g/mol
moles of N2 = 31500 g / 28 g/mol
moles of N2 = 1125 mol
Now, let's compare the moles of each reactant. The reactant that produces the smallest number of moles is the limiting reactant.
In this case, we can see that there are 2610 moles of H2 and 1125 moles of N2.
Since N2 has a smaller number of moles compared to H2, it is the limiting reactant.
Therefore, N2 is the limiting reactant in this reaction.