how many grams of nitrogen trihydride are produced if 621.9 g of nitrogen gas and 134.3 g of hydrogen gas are reacted with each other
To determine the grams of nitrogen trihydride (NH3) produced, we need to find the limiting reactant between nitrogen gas (N2) and hydrogen gas (H2). The balanced chemical equation for the reaction is:
N2 + 3H2 -> 2NH3
Now, let's find the moles of nitrogen gas and hydrogen gas:
Moles of N2 = mass / molar mass
Moles of N2 = 621.9 g / 28.0134 g/mol (molar mass of N2)
Moles of N2 ≈ 22.185 mol
Moles of H2 = mass / molar mass
Moles of H2 = 134.3 g / 2.01588 g/mol (molar mass of H2)
Moles of H2 ≈ 66.646 mol
Now, let's determine the limiting reactant by comparing the moles of N2 and H2. The balanced equation shows that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, the limiting reactant is the one that produces fewer moles of NH3.
Moles of NH3 produced from N2 = (22.185 mol N2)(2 mol NH3 / 1 mol N2) = 44.37 mol NH3
Moles of NH3 produced from H2 = (66.646 mol H2)(2 mol NH3 / 3 mol H2) = 44.431 mol NH3
The moles of NH3 produced from N2 and H2 are similar, indicating that the limiting reactant is the same, regardless of the limiting reactant. Therefore, nitrogen gas (N2) is the limiting reactant.
Now, to calculate the grams of NH3 produced, we will use the stoichiometry of the balanced equation:
Molar mass of NH3 = 17.03052 g/mol (molar mass of NH3)
Grams of NH3 produced = moles of NH3 (from N2) × molar mass of NH3
Grams of NH3 produced = 44.37 mol × 17.03052 g/mol
Grams of NH3 produced ≈ 754.69 g (rounded to the nearest hundredths)
Therefore, approximately 754.69 grams of nitrogen trihydride (NH3) are produced.