How many grams of H2 are needed to produce 13.09 g of NH3?
To determine the amount of H2 needed to produce 13.09 g of NH3, we need to first write out the balanced chemical equation for the reaction:
N2 + 3H2 → 2NH3
From the balanced equation, we can see that 3 moles of H2 are needed to produce 2 moles of NH3.
Next, we need to determine the molar mass of NH3.
Molar mass of NH3 = 14.01 g/mol (N) + 3(1.01 g/mol) = 17.03 g/mol
Now, let's calculate the moles of NH3 produced:
Moles of NH3 = 13.09 g / 17.03 g/mol = 0.767 moles
Given that 2 moles of NH3 require 3 moles of H2, we can set up a proportion to determine the moles of H2 needed to produce 0.767 moles of NH3:
3 moles H2 / 2 moles NH3 = X moles H2 / 0.767 moles NH3
X = 0.767 moles H2 × 3 moles H2 / 2 moles NH3 = 1.15 moles H2
Finally, we can calculate the grams of H2 needed:
Grams of H2 = 1.15 moles H2 × 2.02 g/mol (H2) = 2.32 g H2
Therefore, 2.32 grams of H2 are needed to produce 13.09 g of NH3.