when 4.00 times 10 to the fifth kg of H2 is added to an excess of N2, 1.04 times ten to the sixth kg of NH3 is produced. what is the percentage yield of th ereaction?
To find the percentage yield of the reaction, we need to compare the actual amount of NH3 produced to the maximum amount that could be produced based on stoichiometry.
First, let's write out the balanced chemical equation for the reaction:
N2 + 3H2 -> 2NH3
According to the equation, the stoichiometric ratio is 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Given:
- Mass of H2 = 4.00 x 10^5 kg
- Mass of NH3 produced = 1.04 x 10^6 kg
To find the number of moles of H2 and NH3, we can use their molar masses:
- Molar mass of H2 = 2 g/mol (from periodic table)
- Molar mass of NH3 = 17 g/mol (from periodic table)
Number of moles of H2 = (mass of H2) / (molar mass of H2)
= (4.00 x 10^5 kg) / (2 g/mol)
= 2.00 x 10^5 mol
Number of moles of NH3 = (mass of NH3) / (molar mass of NH3)
= (1.04 x 10^6 kg) / (17 g/mol)
= 6.12 x 10^4 mol
Now let's calculate the maximum moles of NH3 that could be produced based on H2 using the stoichiometric ratio:
Number of moles of NH3 (maximum) = (number of moles of H2) x (stoichiometric ratio of NH3/H2)
= (2.00 x 10^5 mol) x (2 mol NH3 / 3 mol H2)
= 1.33 x 10^5 mol
Finally, we can calculate the percentage yield:
Percentage yield = (actual moles of NH3 produced / maximum moles of NH3 that could be produced) x 100%
= (6.12 x 10^4 mol / 1.33 x 10^5 mol) x 100%
= 46.12%
Therefore, the percentage yield of the reaction is approximately 46.12%.