. How many grams of H2 are required to produce 7.50 grams of ammonia NH3? Use the balanced equation (this time the equation is already balanced):
N_2 (g)+3H_2 (g)⟶2NH_3 (g)
To determine the number of grams of H2 required to produce 7.50 grams of NH3, we need to consider the stoichiometry of the reaction.
From the balanced equation, we can see that 3 moles of H2 are required to produce 2 moles of NH3. This means that the ratio of H2 to NH3 is 3:2.
To convert grams of NH3 to moles, we use the molar mass of NH3:
Molar mass of NH3 = 1(1.008) + 3(1.0079) = 17.031 g/mol
Moles of NH3 = mass of NH3 / molar mass of NH3
Moles of NH3 = 7.50 g / 17.031 g/mol ≈ 0.440 moles of NH3
Now, using the ratio of H2 to NH3, we can calculate the moles of H2 required:
Moles of H2 = (moles of NH3) * (3/2)
Moles of H2 = 0.440 moles * (3/2) = 0.660 moles
To convert moles of H2 to grams, we use the molar mass of H2:
Molar mass of H2 = 2(1.0079) = 2.0158 g/mol
Grams of H2 = moles of H2 * molar mass of H2
Grams of H2 = 0.660 moles * 2.0158 g/mol ≈ 1.322 grams
Therefore, approximately 1.322 grams of H2 are required to produce 7.50 grams of NH3.