Answered

Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

What is the percent yield of this reaction if 22.o g of Mgl2 is produced by the reaction of 25.0 g of Mg with 25.0 g of l2?

Sagot :

ardni313

% yield = 80.719

Further explanation

Given

22.0 g of Mgl₂

25.0 g of Mg

25.0 g of l₂

Required

The percent yield

Solution

Reaction

Mg + I₂⇒ MgI₂

mol Mg = 25 g : 24.305 g/mol = 1.029

mol I₂ = 25 g : 253.809 g/mol = 0.098

Limiting reactant = I₂

Excess reactant = Mg

mol MgI₂ based on I₂, so mol MgI₂ = 0.098

Mass MgI₂ (theoretical):

= mol x MW

= 0.098 x 278.114

= 27.255 g

% yield = (actual/theoretical) x 100%

% yield = (22 / 27.255) x 100%

% yield = 80.719

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.