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

Answer:

$m_{NH_3}^{leftover}=57.88g$

$Y= 92.0\%$

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:

$n_{NI_3}^{by\ IBr}=164.8gIBr*\frac{1molIBr}{206.81gIBr}*\frac{1molNI_3}{3molIBr} =0.266molNI_3\\\\n_{NI_3}^{by\ NH_3}=62.4gNH_3*\frac{1molNH_3}{17.03gNH_3}*\frac{1molNI_3}{1molNH_3} =3.66molNI_3$

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:

$m_{NH_3}^{reacted}=0.266molNI_3*\frac{1molNH_3}{1molNI_3}*\frac{17.03gNH_3}{1molNH_3}=4.52gNH_3$

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:

$m_{NI_3}^{theoretical}=0.266molNI_3*\frac{394.72gNI_3}{1molNI_3} =104.8gNI_3$

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

$Y=\frac{96.4g}{104.8g}*100\%\\\\Y= 92.0\%$

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