0.498 moles of copper(II) phthalocyanine would be produced by
the complete cyclotetramerization of 255 grams of phthalonitrile in the
presence of excess copper(ll) chloride.
Copper(ll) phthalocyanine (Cu(C₃₂H₁₆N₈)) is produced by the cyclotetramerization of phthalonitrile (C₈H₄N₂) according to the following reaction: 4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g) How many moles of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 255 grams of phthalonitrile in the presence of excess copper(II) chloride?
Let's consider the following balanced equation.
4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g)
The molar mass of C₈H₄N₂ is 128.13 g/mol. The moles corresponding to 255 g of C₈H₄N₂ are:
[tex]255 g \times \frac{1mol}{128.13 g} = 1.99 mol[/tex]
The molar ratio of C₈H₄N₂ to Cu(C₃₂H₁₆N₈) is 4:1. The moles of Cu(C₃₂H₁₆N₈) produced from 1.99 moles of C₈H₄N₂ are:
[tex]1.99 mol C_8H_4N_2 \times \frac{1molCu(C_{32}H_{16}N_8)}{4mol C_8H_4N_2} = 0.498 mol Cu(C_{32}H_{16}N_8)[/tex]
0.498 moles of copper(II) phthalocyanine would be produced by
the complete cyclotetramerization of 255 grams of phthalonitrile in the
presence of excess copper(ll) chloride.
You can learn more about stoichiometry here: https://brainly.com/question/22288091
