Question

Question 7

Answer 1

Answer:

84.3 g of nitrogen triiodide is the theoretical yield.

Explanation:

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In this case, according to the chemical reaction:

$N_2 + 3I_2 \rightarrow 2NI_3$

It is possible to compute the theoretical yield of nitrogen triiodide by each reactant via stoichiometry as shown below:

$m_{NI_3}^{by\ N_2}=4.49gN_2*\frac{1molN_2}{28.01gN_2} *\frac{2molNI_3}{1molN_2}*\frac{394.72gNI_3}{1molNI_3} =126.55gNI_3\\\\m_{NI_3}^{by\ I_2}=81.3gI_2*\frac{1molI_2}{253.81gI_2} *\frac{2molNI_3}{3molI_2}*\frac{394.72gNI_3}{1molNI_3} =84.29gNI_3$

Therefore, we infer that the smallest amount is the correct theoretical yield as it comes from the limiting reactant, in this case, diatomic iodine as it yields 84.3 g (three significant figures) of nitrogen triiodide as the theoretical yield; incidentally, nitrogen acts as the excess reactant.

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