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Answer:
[tex]m_{NH_3}^{leftover}=57.88g[/tex]
[tex]Y= 92.0\%[/tex]
Explanation:
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In this case, according to the following chemical reaction between iodine monobromide and ammonia:
[tex]3IBr+NH_3\rightarrow NI_3+3HBr[/tex]
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:
[tex]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[/tex]
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:
[tex]m_{NH_3}^{reacted}=0.266molNI_3*\frac{1molNH_3}{1molNI_3}*\frac{17.03gNH_3}{1molNH_3}=4.52gNH_3[/tex]
And therefore the leftover of ammonia is:
[tex]m_{NH_3}^{leftover}=62.4g-4.52g=57.88g[/tex]
Next, the percent yield is calculated by firstly calculating the theoretical yield of nitrogen triiodide as follows:
[tex]m_{NI_3}^{theoretical}=0.266molNI_3*\frac{394.72gNI_3}{1molNI_3} =104.8gNI_3[/tex]
And finally the percent yield by dividing the given actual yield of 96.4 g by the previously computed theoretical yield:
[tex]Y=\frac{96.4g}{104.8g}*100\%\\\\Y= 92.0\%[/tex]
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