If 250 gN, and 18.75 g H, are used for this reaction How much of the excess reactant will be left over?
To determine how much of the excess reactant will be left over, we first need to determine the limiting reactant in the reaction.
Let's first convert the mass of each reactant to moles:
1. Convert 250 g of nitrogen (N2) to moles:
molar mass of N2 = 14.01 g/mol
moles of N2 = 250 g / 28.02 g/mol = 8.92 moles
2. Convert 18.75 g of hydrogen (H2) to moles:
molar mass of H2 = 1.01 g/mol
moles of H2 = 18.75 g / 2.02 g/mol = 9.30 moles
Now we need to determine the limiting reactant by comparing the moles of each reactant to the stoichiometric ratio of the reaction:
N2 + 3H2 -> 2NH3
From the balanced equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Let's calculate the moles of NH3 that can be produced by each reactant:
For N2: 8.92 moles N2 * (2 moles NH3 / 1 mole N2) = 17.84 moles NH3
For H2: 9.30 moles H2 * (2 moles NH3 / 3 moles H2) = 6.20 moles NH3
Since H2 produces less NH3, it is the limiting reactant in this case.
Now, let's calculate the amount of excess reactant left over:
Moles of H2 reacted = 9.30 moles - 6.20 moles = 3.10 moles
Mass of excess H2 left over = 3.10 moles * 2.02 g/mol = 6.26 g
Therefore, 6.26 g of excess hydrogen will be left over after the reaction.