Answer:
9.83 grams
Step-by-step explanation:
The balanced chemical equation for the reaction between hydrogen gas (H2) and nitrogen gas (N2) to form ammonia (NH3) is:
N2 + 3 H2 → 2 NH3
From the balanced equation, we can see that the stoichiometric ratio of H2 to NH3 is 3:2. This means that for every 3 moles of H2 used, 2 moles of NH3 are produced.
To determine the amount of NH3 that can be produced from 3.50 grams of H2, we first need to convert the mass of H2 to moles using its molar mass:
Molar mass of H2 = 2.02 g/mol (atomic mass of H) x 2 = 4.04 g/mol
Moles of H2 = Mass of H2 / Molar mass of H2
Moles of H2 = 3.50 g / 4.04 g/mol
Moles of H2 = 0.866 moles
Using the mole ratio between H2 and NH3, we can calculate the theoretical yield of NH3:
N2 + 3 H2 → 2 NH3
3 mol H2 / 2 mol NH3 = 0.866 mol H2 / x mol NH3
x = (2/3) x 0.866 = 0.577 moles NH3
Therefore, the theoretical yield of NH3 is 0.577 moles.
To convert this to grams of NH3, we can use its molar mass:
Molar mass of NH3 = 14.01 g/mol (atomic mass of N) + 3(1.01 g/mol) (atomic mass of H) = 17.03 g/mol
Mass of NH3 = Moles of NH3 x Molar mass of NH3
Mass of NH3 = 0.577 mol x 17.03 g/mol
Mass of NH3 = 9.83 g (rounded to two decimal places)
Therefore, if 3.50 grams of H2 react, the maximum mass of NH3 that can be produced is 9.83 grams.