Respuesta :

kenmyna
 Theoretical  yield in grams for this  reaction  under given condition  is  6.341 grams

calculation

write the  reacting equation
 3 H2 + N2 →  2NH3
find the  moles  of  each reactant
moles  =mass/molar mass
  moles  of H2 =  1.12g/2 g/mol =  0.56 moles
  moles  of  N2 = 9.72 g/28 g/mol = 0.341  moles

by  use of mole ratio between H2 to NH3 which is   3: 2 the moles   of NH3 =  0.56 x 2/3=  0.373  moles

by use  of mole ratio  between N2  to  NH3  which  is 1: 2 the  moles of  NH3 = 0.341  x2/1   =  0.682  moles

this relationship   show that  H2   is the limiting reagent   hence the moles of NH3  = 0.373 moles

The theoretical mass  =  moles of NH3  x molar mass of NH3
molar mass  of NH3 =  14+ 3 = 17 g/mol

theoretical mass is therefore =   17 g/mol   x 0.373  moles =6.341  grams


mariejoepagaduan

H₂ is the limiting reactant and the theoretical yield is 6.31 g of NH₃.

Further Explanation

In order to determine the theoretical yield, the following steps must be done:

  1. Write the balanced chemical equation.
  2. Determine the limiting reactant. This is the reactant that will determine the amount of NH₃ that will actually form.
  3. Determine the theoretical yield for NH₃ when the limiting reactant is used.

STEP 1. The balanced chemical equation for the given reaction is:

3 H₂ + N₂ → 2 NH₃

STEP 2. Determining the Limiting Reactant

The Limiting Reactant (LR) will produce fewer moles of the products. To check  which of the reactants H₂ or N₂ is the LR ,do dimensional analysis:

  • For H₂:

[tex]mol \ of \ NH_3 \ produced = 1.12 \ g \ H_2 \times (\frac{1 \ mol H_2}{2.016 \ g \ H_2}) (\frac{2 \ mol \ NH_3}{3 \ mol \ H_2})\\ \\\boxed {mol \ of \ NH_3 \ produced = 0.3704 \ mol}[/tex]

  • For N₂:

[tex]mol \ of \ NH_3 \ produced = 9.72 \ g \ N_2 \times (\frac{1\ mol \ N_2}{28.014 \ g \ N_2}) (\frac{2 \ mol \ NH_3}{1 \ mol \ N_2})\\\boxed {mol \ of \ NH_3 \ produced = 0.6939 \ mol}[/tex]

Since H₂ produces fewer moles of NH₃, then it is the limiting reactant. Its given amount will be used to determine the theoretical yield.

STEP 3. Determining the Theoretical Yield

From Step 2 it is known that 0.3704 moles of NH₃ will be produced. This needs to be converted to grams.

[tex]mass \ of \ NH_3 \ produced = 0.3704 \ mol \ NH_3 \times (\frac{17.031 \ g \ NH_3}{1 \ mol \ NH_3})\\mass \ of \ NH_3 \ = 6.308 \ g\\\boxed {\boxed {mass \ of \ NH_3 \ produced = 6.31 \ g}}[/tex]

Since the answer only requires 3 significant figures, the final answer is 6.31 grams NH₃.

Learn More  

  1. Learn More about Limiting Reactant brainly.com/question/7144022
  2. Learn More about Excess Reactant brainly.com/question/6091457  
  3. Learn More about Stoichiometry brainly.com/question/9743981

Keywords: stoichiometry, theoretical yield

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