How many grams of NH3 can be produced
from the reaction of 17.8 moles of H2 and a sufficient supply of N2?
N2 + 3H2 ! 2NH3
To determine the grams of NH3 produced, we need to use the balanced chemical equation and the molar mass of NH3.
First, let's calculate the molar mass of NH3:
NH3 = 1(N) + 3(H) = 14.01 g/mol + 3(1.01 g/mol) = 17.03 g/mol
Next, we need to find the moles of NH3 produced by using the stoichiometry of the balanced equation:
From the balanced equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Therefore, we can set up a proportion to find the moles of NH3 produced:
1 mol N2 / 3 mol H2 = x mol NH3 / 17.8 mol H2
Now, solve for x:
x = (1 mol N2 / 3 mol H2) * (17.8 mol H2) = 5.93 mol NH3
Finally, let's calculate the grams of NH3 produced:
Grams of NH3 = moles of NH3 * molar mass of NH3
Grams of NH3 = 5.93 mol NH3 * 17.03 g/mol = 100.93 g of NH3
Therefore, 100.93 grams of NH3 can be produced from the reaction of 17.8 moles of H2 and a sufficient supply of N2.