How many moles of nitrogen do you need to make 78 g of ammonia?
The reaction is
N2 + 3H2 = 2 NH3
78 moles of NH3 is 4 moles, since the molecular weight is 17.
You need half as many moles of N2 as you make of NH3.
5.20
To determine how many moles of nitrogen (N2) are needed to make 78 grams of ammonia (NH3), we need to follow a few steps.
Step 1: Determine the molar mass of ammonia (NH3) and nitrogen (N2).
The molar mass of NH3 is calculated by adding up the atomic masses of its constituent elements:
Molar mass of N = 14.01 g/mol (nitrogen)
Molar mass of H = 1.008 g/mol (hydrogen)
So, molar mass of NH3 = 14.01 g/mol + (3 * 1.008 g/mol) = 17.03 g/mol
The molar mass of N2 is calculated by multiplying the atomic mass of nitrogen (N) by 2:
Molar mass of N2 = 14.01 g/mol * 2 = 28.02 g/mol
Step 2: Calculate the number of moles of nitrogen (N2).
Using the given mass of ammonia (NH3) and its molar mass, we can calculate the moles of ammonia:
Moles of NH3 = Mass of NH3 / Molar mass of NH3
Moles of NH3 = 78 g / 17.03 g/mol ≈ 4.58 mol
Since ammonia has one mole of nitrogen (N2) for every mole of ammonia, the moles of nitrogen are also equal to 4.58 mol.
Therefore, you would need approximately 4.58 moles of nitrogen (N2) to produce 78 grams of ammonia (NH3).