If 250 g N2, and 18.75 g H2, are used for this reaction, How much of the excess reactant will be left over?
To determine the excess reactant left over, we first need to calculate the moles of each reactant using their respective molar masses.
1. Calculate the moles of N2:
Molar mass of N2 = 28.02 g/mol
Moles of N2 = 250 g / 28.02 g/mol = 8.92 mol
2. Calculate the moles of H2:
Molar mass of H2 = 2.02 g/mol
Moles of H2 = 18.75 g / 2.02 g/mol = 9.30 mol
Next, we need to determine the limiting reactant in the reaction. The balanced chemical equation for the reaction is:
N2 + 3H2 -> 2NH3
From the equation, we can see that the ratio of moles of N2 to H2 is 1:3. This means that N2 is the limiting reactant.
Since N2 is the limiting reactant, all of the N2 will react completely and only some of the H2 will react. Now we need to calculate the amount of H2 that will react with the given amount of N2:
Moles of N2 : Moles of H2 = 1:3
Moles of H2 reacting with N2 = 8.92 mol N2 x (3 mol H2 / 1 mol N2) = 26.76 mol H2
Since we only have 9.30 mol of H2, it is the excess reactant. The amount of excess reactant left over can be calculated by subtracting the moles of H2 that reacted with N2 from the total moles of H2:
Excess H2 left over = 9.30 mol H2 - 26.76 mol H2 = -17.46 mol H2
Since we cannot have a negative amount of a substance in a chemical reaction, the excess reactant left over is 0 mol.