1.47 g H 2 is allowed to react with 9.71 g N 2 , producing 1.31 g N H 3 . Part A

Question

1.47g H2 is allowed to react with 9.71g N2, producing 1.31g NH3. Part A What is the theoretical yield for this reaction under the given conditions?

Part B What is the percent yield for this reaction under the given conditions? 1.47g H2 is allowed to react with 9.71g N2, producing 1.31g NH3. 1.47g H2 is allowed to react with 9.71g N2, producing 1.31g NH3. Part A What is the theoretical yield for this reaction under the given conditions?

Part B What is the percent yield for this reaction under the given conditions? Part A What is the theoretical yield for this reaction under the given conditions?

Part B What is the percent yield for this reaction under the given conditions? What is the percent yield for this reaction under the given conditions?

Explanation / Answer

N2 + 3H2 ----> 2 NH3

as per balanced reaction

1 mole of N2 reacts with 3 mol of H2 to form 2 mol of NH3

now

1.47g H2 is allowed to react with 9.71g N2, producing 1.31g NH3.

or we convert grams of each given to moles using molar mass of each

1.47/2.016 = 0.73 moles of H2 reacts with 9.71/28 = 0.35 moles of N2

now we look for limiting reactant

asper reaction 1 mol of N2 reacts with 3 moles of H2

so 0.35 moles of N2 (given) will need 3* 0.35 = 1.05 moles of H2

so we require 1.05 moloes of H2 but we have only 0.73 moles given.It means moles of H2 are

less than needed and will be used up completely.So H2 will be limiting reactant.

Now

3 moles of H2 gives 2 moles NH3

0.73 moles of H2 will give =( 2 moles NH3/ 3 moles H2)* 0.73 moles H2

= 0.487 moles NH3

Therefore theoretical yield of NH3 = 0.487 moles = 0.487*17.024= 8.29 grams NH3 (answer)

b)% yield = (actual yield/Theoretical yield) *100 as per reaction

% yield =

(1.31/8.29) *100 = 15.8 % (answer)

s

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