How many grams of NH3 can be produced form the reaction of 28 g of nitrogen and 25g of hydrogen?
I doubt if N2 and H2 would react to make NH3, however,
because Nitrogen is 28/(28+3) or 90 percent of NH3, so Hydrogen is then 3 percent, it is obvious that most of the mass is in N.
So if you use 28 grams of Nitrogen, you only need a few grams of Hydrogen
grams. If 28 grams is 97 percent, the the hydrogen 3 percent must be...
28/97=x/3 or x= .9 grams, so how much NH3? what is 28+.9*3 ?
At the right temperature and pressure and with a suitable catalyst, this reaction produces NH3 commercially. This is a limiting reagent (LR) problem.
N2 + 3H2 ==> 2NH3
mols N2 = 28 g/28 g/mol = 1 mole. You will need 3 mols H2 and you have well over that (25/2 = 12.5 mols H2 initially) so N2 is the LR.
That will produce 1 mol N2 x (2 mols NH3/1 mol N2) = 2 mols NH3.
g NH3 produced = 2 mols x molar mass NH3 = 34 grams.
To determine the number of grams of NH3 produced, we need to first balance the chemical equation for the reaction:
N2 + 3H2 -> 2NH3
According to the balanced equation, 1 mole of nitrogen (N2) reacts with 3 moles of hydrogen (H2) to produce 2 moles of ammonia (NH3).
Step 1: Convert the given masses of nitrogen and hydrogen to moles.
- Moles of nitrogen (N2) = 28 g / molar mass of nitrogen (N2)
- Moles of hydrogen (H2) = 25 g / molar mass of hydrogen (H2)
The molar mass of nitrogen (N2) is approximately 28 g/mol, and the molar mass of hydrogen (H2) is approximately 2 g/mol.
Moles of nitrogen (N2) = 28 g / 28 g/mol = 1 mole
Moles of hydrogen (H2) = 25 g / 2 g/mol = 12.5 moles
Step 2: Determine the limiting reactant.
To identify the limiting reactant, we compare the moles of nitrogen and hydrogen, keeping in mind the stoichiometry of the balanced equation.
According to the balanced equation, 1 mole of nitrogen (N2) reacts with 3 moles of hydrogen (H2). Thus, we would need 3 moles of hydrogen for every mole of nitrogen.
Since we have only 1 mole of nitrogen and 12.5 moles of hydrogen, the nitrogen is the limiting reactant.
Step 3: Calculate the moles of ammonia (NH3) based on the limiting reactant.
According to the balanced equation, 1 mole of nitrogen (N2) reacts to produce 2 moles of ammonia (NH3).
Moles of ammonia (NH3) = 2 moles of NH3 per 1 mole of N2 * 1 mole of N2
Moles of ammonia (NH3) = 2 moles * 1 mole = 2 moles
Step 4: Convert moles of ammonia to grams.
Since the molar mass of ammonia (NH3) is approximately 17 g/mol, we can convert moles to grams.
Grams of ammonia (NH3) = 2 moles * 17 g/mol = 34 grams
Therefore, 34 grams of NH3 can be produced from the reaction of 28 g of nitrogen and 25 g of hydrogen.