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In this reaction: Mg (s) + I₂ (s) → MgI₂ (s), if 10.0 g of Mg reacts with 60.0 g of I₂, and 57.84 g of MgI₂ form, what is the percent yield?

Answer :

Leora03

Answer:

88.1% (to 3 s.f.)

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