Question

If 164.8 g of iodine monobromide is combined with 62.4 g of ammonia, determine the amount of excess reactant that remains after the reaction is completed and the percentage yield of nitrogen triiodide if 96.4 is produced?

Answer 1

`m_(NH_3)^(leftover)=57.88g`

`Y= 92.0\%`

Step-by-step explanation:

Hello there!

In this case, according to the following chemical reaction between iodine monobromide and ammonia:

`3IBr+NH_3\rightarrow NI_3+3HBr`

It turns out firstly necessary to identify the limiting reactant, by considering the proper molar masses and the 3:1 and 1:1 mole ratios of iodine monobromide to nitrogen triiodide and ammonia to nitrogen triiodide respectively:

`n_(NI_3)^(by\ IBr)=164.8gIBr*(1molIBr)/(206.81gIBr)*(1molNI_3)/(3molIBr) =0.266molNI_3\\\\n_(NI_3)^(by\ NH_3)=62.4gNH_3*(1molNH_3)/(17.03gNH_3)*(1molNI_3)/(1molNH_3) =3.66molNI_3`

Thus, we conclude that the limiting reactant is IBr as is yields the fewest moles of nitrogen triiodide product. Next, we can calculate the reacted grams of ammonia as the excess reactant:

`m_(NH_3)^(reacted)=0.266molNI_3*(1molNH_3)/(1molNI_3)*(17.03gNH_3)/(1molNH_3)=4.52gNH_3`

And therefore the leftover of ammonia is:

`m_(NH_3)^(leftover)=62.4g-4.52g=57.88g`

Next, the percent yield is calculated by firstly calculating the theoretical yield of nitrogen triiodide as follows:

`m_(NI_3)^(theoretical)=0.266molNI_3*(394.72gNI_3)/(1molNI_3) =104.8gNI_3`

And finally the percent yield by dividing the given actual yield of 96.4 g by the previously computed theoretical yield:

`Y=(96.4g)/(104.8g)*100\%\\\\Y= 92.0\%`

Best regards!