Answered

Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our 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

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.