Question

1.91 g H2 is allowed to react with 9.86 g N2, producing 1.74 g NH3. 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.

Answer 1

`"Theoretical yield"=32.5g NH_3`

Explanation:

1. Write and balance the equation

   `3H_2(g)+N_2(g)->2NH_3(g)`

2. Find the molar masses of the reactants and the product. Values are obtainable from the periodic table.

   `H_2=2"g/mol"`
   `N_2=28"g/mol"`
   `NH_3=17"g/mol"`

3. Given the masses of the reactants, the number of moles can be calculated as illustrated below.

   `etaH_2=1.91cancel(gH_2)xx(1molH_2)/(2cancel(gH_2))=0.955molH_2`
   `etaN_2=9.86cancel(gN_2)xx(1molN_2)/(28cancel(gN_2))=0.352molN_2`

4. Now, identifying how much each reactant requires if the reaction goes into completion. Refer to the balanced equation for the needed molar ratio.
   a. `ul(etaH_2=0.955mol " available"):`
   `etaN_2=0.955cancel(molH_2)xx(1molN_2)/(3cancel(molH_2))=0.318molN_2`
   Thus;
   `0.955molH_2-=0.318molN_2`
   b. `ul(etaN_2=0.352mol " available":`
   `etaH_2=0.352cancel(molN_2)xx(3molH_2)/(1cancel(molN_2))=1.056molH_2`
   Thus;
   `0.352molN_2-=1.056molH_2`

5. Find the limiting reactant through comparison: the required and the available number of moles; i.e.,
   `(H_2" available")/(0.955molH_2)<(H_2" required")/(1.056molH_2)`
   `color(red)(H_2 " is the limiting reactant."`
   `(N_2" available")/(0.352molN_2)>(N_2" required")/(0.318molN_2)`
   `N_2 " is the excess reactant."`

6. Knowing the limiting reactant, the theoretical yield `(TY)` can be computed as shown below. Refer to the balanced equation for the mole ratio and the molar mass of the `NH_3` as reflected in `"step" 2` for the following series of conversions:
   `=0.955cancel(molH_2)xx(2molNH_3)/(1cancel(molH_2))`
   `=1.91cancel(molNH_3)xx(17gNH_3)/(1cancel(molNH_3))`
   `=32.5gNH_3`