Question

Question #bf4d3

Answer 1

`41%`

Explanation:

First, we need to calculate theoretical yield from the balanced equation.

`4NH_3 + 6NO -> 5N_2 + 6H_2O` this shows that we should get 5 moles of `N_2` for every 4 moles of `NH_3` reacted.
10.0g of `NH_3` is `(10.0/17) = 0.588` moles. The resulting `N_2` should then be `5/4 * 0.588 = 0.735` moles. From this value we can either calculate grams, for a mass/mass % yield, or convert our given grams into moles for a mole/mole (volume/volume) % yield.

Mass basis: `0.735"mol" N_2 * (28g N_2)/(mol N_2) = 20.58` g `N_2` Theoretical `(8.50/20.58 * 100) = 41.3%` by mass

Mole basis: `(8.50g N_2)/((28g N_2)/(mol N_2)) = 0.304` moles `N_2` `(0.304/0.735 * 100) = 41.4%` by volume