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Sagot :
Maximum mass of p2i4 that can be prepared is 10.33 gm.
The balanced chemical reaction would be as follows:
5P4O6 + 8I2 → 4P2I4 + 3P4O10
Molar Mass of P4O6 = 219.88 g/mol
Molar mass of molecular iodine = 253.8 g/mol
Molar mass of P2I4 = 569.57 g/mol
We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:
8.08g of P4O6 = 8.08/219.88 = 0.036 Moles
9.21 g of iodine = 9.21/253.8 = 0.03627 moles
Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:
0.03627 Moles of iodine will react with (0.03627 x 5/8) 0.02267 moles of P4O6 and will give (0.02267 x4/5) 0.181 moles of P2I4.
That is equal to (0.0181 x 569.57) 10.33 gm of P2I4.
To learn more about Mole Concept visit:
https://brainly.com/question/2162369
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