how many grams of NH3 can be produced from 4.35moles of N2 and excess H2
3H2 + N2 ==> 2NH3
4.35 mols N2 x ((2 mols NH3/1 mol N2) = 4.35 x 2 = ?
g NH3 = mols NH3 x molar mass NH3
To determine the number of grams of NH3 that can be produced from 4.35 moles of N2 and excess H2, we need to follow a series of steps.
Step 1: Write a balanced chemical equation for the reaction. The balanced equation for the reaction of N2 and H2 to produce NH3 is:
N2 + 3H2 -> 2NH3
Step 2: Determine the molar ratio between N2 and NH3 from the balanced equation. From the balanced equation, we can see that it takes 1 mole of N2 to produce 2 moles of NH3.
Step 3: Calculate the number of moles of NH3 that can be produced. Since we have 4.35 moles of N2, we can use the molar ratio to calculate the moles of NH3:
4.35 moles N2 * (2 moles NH3 / 1 mole N2) = 8.70 moles NH3
Step 4: Convert the moles of NH3 to grams. To do this, we need to use the molar mass of NH3, which is 17.03 grams/mole.
8.70 moles NH3 * 17.03 grams/mole = 148.83 grams NH3
Therefore, 148.83 grams of NH3 can be produced from 4.35 moles of N2 and excess H2.