Of the following elements, which one would have the smallest ionization energy ?
1 answer:
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Answer:
6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.
Explanation:
We are given the chemical equation:
And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.
Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:
- The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)
- The ratio between NH₃ and Cu is 2:3.
- The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)
Dimensional Analysis:
- The amount of N₂ produced:
- The amount of Cu produced:
- And the amount of H₂O produced:
In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.
In order of increasing percent water content:CoCl₂.6H₂O, Ba(OH)₂.8H₂O, MgSO₄.7H₂O
<h3>Further explanation</h3>CoCl₂.6H₂O.MW=237.90 g/mol
6H₂O MW = 6.18=108 g/mol
MgSO₄.7H₂O.MW=246.48 g/mol
MW 7H₂O = 7.18=126 g/mol
Ba(OH)₂.8H₂O MW=315.48 g/mol
MW 8H₂O = 8.18=144 g/mol