Question

Question #89a6e

Answer 1

`a. N_2(g)+3H_2(g)->2NH_3(g)`
`b. ~~577gNH_3`
`c. ~~69.0% yield`

Explanation:

1. Write and balance the given chemical equation.
   `color(red)(N_2+3H_2.->2NH_3`

2. Find the required molar masses of the involved substances.
   `N_2=28.0" g/mol"`
   `NH_3=17.0" g/mol"`

3. Given the masses of both raw material and the production output, theoretical yield of this production run can be computed as follows:
   `=475cancel(gN_2)xx(1molN_2)/(28.0cancel(gN_2))`
   `=16.96molN_2~~17.0molN_2`

Now, convert mole `O_2` `(etaO_2)` to `etaNH_3` using the molar ratio of these two(2) substances that is obtainable from the balanced equation described in `ul ("step " 1)`.
`=17.0cancel(molN_2)xx(2molNH_3)/(1cancel(molN_2))`
`=33.93molNH_3~~34.0molNH_3`

Then, find the `"theoretical yield"`. Knowing the molar mass of `NH_3`, conversion factor is obtainable and the production output can be computed. Make sure units work out and the desired unit is attained; i.e.,
`color(red)(=34.0cancel((molNH_3)xx(17.0gNH_3)/(1cancel(molNH_3))`
`color(red)(=576.79gNH_3~~577gNH_3)`

1. Finally, find the percent yield in this production run; i.e.,
   `% yield=("actual yield")/("theoretical yield")xx100`
   `% yield=(397cancel(gNH_3))/(577cancel(gNH_3))xx100`
   `color(red)(% yield=68.8%~~69.0%`