Question

In this reaction: Mg (s) + I₂ (s) → MgI₂ (s), if 10.0 g of Mg reacts with 60.0 g of I₂, and 53.88 g of MgI₂ form, what is the percent yield?

Answer 1

• Mass of reactants=10+60=70.0g
• Mass of MgI_2=53.88g
• Yield=70-53.88=16.12g

Percentage

$\\ \tt\hookrightarrow \dfrac{16.12}{70}\times 100=23.02\%$

Yield percentage is 23.02%

Answer 2

Answer:

I don't know please very sorry for your birthday