yehia6
Answered

At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

What is the percentage of the actual yield of the reaction of 2.5 mol of Fe(NO3),
with 3.6 mol of Na2Co3, to form 6.3 mol of NaNO3 according to the reaction :
2Fe(NO3)3 + 3NaC03 - >
Fe2(CO3)3 + 6NaNO3 ?​

Sagot :

Neetoo

Answer:

Percent yield = 87.5%

Explanation:

Given data:

Number of moles of Fe(NO₃)₃ = 2.5 mol

Number of moles of Na₂CO₃ = 3.6 mol

Actual yield of NaNO₃ = 6.3 mol

Percent yield of NaNO₃ = ?

Solution:

Chemical equation:

2Fe(NO₃)₃ + 3Na₂CO₃        →     Fe₂(CO₃)₃ + 6NaNO₃

now we will compare the moles of both reactant with NaNO₃.

                   Fe(NO₃)₃          :              NaNO₃

                       2                   :                  6

                     2.5                  :            6/2×2.5 = 7.5

                  Na₂CO₃             :              NaNO₃

                       3                   :                  6

                     3.6                  :            6/3×3.6 = 7.2

Theoretical yield of NaNO₃ is 7.2 mol.

Percent yield:

Percent yield = actual yield / theoretical yield ×100

Percent yield = 6.3 mol / 7.2mol ×100

Percent yield = 87.5%

Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.