Question

Consider the following reaction. 5NaN_3(s) → NaNO_3(aq) → 3Na_2O(s) + 8N_2(g). If 2.50 g of NaN_3(s) react with excess NaNO_3aq), 0.995 g of N_3(g) is recovered. What is the percent yield?

a) 35.6%
b) 42.2%
c) 53.6%
d) 64.8%

Answer 1

To calculate the percent yield, the actual yield of 0.995 g of N2 is compared to the theoretical yield, calculated from the stoichiometry of the reaction. The calculated percent yield is 57.7%, which does not match any of the provided answer choices.

The right answer is c) 53.6%

Step-by-step explanation:

To calculate the percent yield of a chemical reaction, we need to compare the actual yield to the theoretical yield. The reaction in question is the decomposition of sodium azide (NaN3), which we can write as:

5NaN3(s) → 3Na bO(s) + 8N2(g).

First, we will calculate the theoretical yield of N2 based on the provided mass of NaN3. The molar mass of NaN3 is approximately 65 g/mol and that of N2 is approximately 28 g/mol. Using stoichiometry, we can determine the number of moles of N2 produced from the given 2.50 g of NaN3.

(2.50 g NaN3) / (65 g/mol) x (8 moles N2 / 5 moles NaN3) = 0.061538462 moles of N2 which is theoretically produced.
The theoretical yield in grams of N2 is then 0.061538462 moles x 28 g/mol = 1.72307694 g.

To calculate the percent yield, we use the formula:

Percent yield = (Actual yield / Theoretical yield) x 100%
Percent yield = (0.995 g / 1.723 g) x 100% = 57.7%

The closest answer choice to this calculated percent yield is not listed in the provided options, so there may be an error in the question or in the available answer choices.

The right answer is c) 53.6%