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

Answer:

The story is a/an

Explanation:

please tell me answer

Answer 2

We know the law of conservation of mass

• It states that mass is neither formed nor destroyed in any chemical reaction.
• Mass of reactants=Mass of products.

Here

• Mg and I_2 are reactants
• MgI_2 is product with some yield.
• Mass of reactants=10+60.0=70.0g
• Mass of MgI_2=53.88g
• Mass of yield=Product-MgI_2=70-53.88=16.12g

Lets find the percentage

$\\ \tt\hookrightarrow \dfrac{Mass\:of\:yield}{Total\:mass}\times 100$

$\\ \tt\hookrightarrow \dfrac{16.12}{70}\times 100$

$\\ \tt\hookrightarrow 0.23028(010)=23.028\%$