m/myct/itemView?assignmentProblemID-97505274 K12 of 19 le 1.41 g H2 is allowed t

Question

m/myct/itemView?assignmentProblemID-97505274 K12 of 19 le 1.41 g H2 is allowed to react with 9.54 g N2. producing 1.81 g NH3. Part A What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units. View Available Hint(s) 7.44 g Submit X Incorrect: Try Again; 5 attempts remaining Part B What is the percent yield for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units. View Available Hint(s) 74.0 % Submit Previous Answers

Explanation / Answer

Answer:-

Therefore the balance chemical reaction is,

3 H2 + N2 2 NH3

First we calculate the number of moles

For H2

(1.41 g H2) / (2.014 g/mol H2) = 0.7001 mol H2

For N2

(9.54 g N2) / (28.012 g/mol N2) = 0.3406 mol N2

0.7001 mole of H2 would react completely with 0.7001 x(1/3) =0.2334 mole of N2, but there is more N2 present than that so N2 is in excess and H2 is the limiting reactant.

Part A:-

Therefore the formula is,

Grams product = grams reactant x (mole ratio product/reactant) x (molar mass of product/1 mol product)

(0.7001 mol H2) x (2 mol NH3 / 3 mol H2) x (17.0.27 g NH3/mol) = 7.95 g NH3 in theoretical yield

Part B:-

Therefore the formula is,

% yield = (actual yield / theoretical yield) × 100%

              = (1.81 / 7.95) × 100%

              = 22.77%

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