To answer this question Kc of this reaction OR the conc of reactants nd products at equilibria is required.
Explanation:
Burning and other changes in matter do not destroy matter. The mass of matter is always the same before and after the changes occur. The law of conservation of mass states that matter cannot be created or destroyed
I’m not sure what this one would be
Answer:
82.08 %
Explanation:
- <em>The percent yield of the reaction = [(actual yield)/(calculated yield)] x 100.
</em>
Actual yield = 26.80 kg.
- <em><u>To get the calculated yield:
</u></em>
- The balanced equation of reacting N2 with H2 to produce NH3 is:
N₂ + 3H₂ → 2NH₃
- It is clear that 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
- N₂ is present in excess and H₂ is the limiting reactant.
- We need to convert the mass of H₂ added (5.79 kg) to moles using the relation:
n = mass /molar mass = (5790 g) / (2.01 g/mol) = 2880.6 mol.
- We can get the no. of moles of NH₃ produced.
<u><em>Using cross multiplication:
</em></u>
3.0 mole of H₂ produce → 2.0 moles off NH₃, from the stichiometry.
2880.6 mol of H₂ produces → ??? moles of NH₃.
- The no. of moles of NH₃ produced = (2880.6 mol)(2.0 mol) / (3.0 mol) = 1920.4 mol.
- We can know get the calculated yield of NH₃ = no. of moles x molar mass = (1920.4 mol) (17.00 g/mol) = 32646.76 g ≅ 32.65 kg.
∴ <em>The percent yield of the reaction = [(actual yield)/(calculated yield)] x 100 = </em>[(26.8 kg) / (32.65 kg)] x 100 <em>= 82.08 %.</em>
