Nitroglycerin (C₃H₅N₃O₉) is a powerful explosive. Its decomposition may be represented by

 4C₃H₅N₃O₉ → 6N₂ + 12CO₂ + 10H₂O + O₂

This reaction generates a large amount of heat and gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion.

 (a) What is the maximum amount of O₂ in grams that can be obtained from 2.50 × 10² g of nitroglycerin?

(b) Calculate the percent yield in this reaction if the amount of O₂ generated is found to be 7.84 g.

a) Maximum amount of O₂

Given:

Mass of C₃H₅N₃O₉ = 2.50 × 10² g 

Molar mass of C₃H₅N₃O₉ = 3(C) + 5(H) + 3(N) + 9(O) = 36 g/mol + 5 g/mol + 42 g/mol + 144 g/mol = 227 g/mol

Convert mass of C₃H₅N₃O₉ into moles as follows:

Moles of C₃H₅N₃O₉ = Mass/Molar mass = 2.50 × 10² g/227 g/mol = 1.1013 mol

The equation that represents the reaction that leads to the production of O₂ has been given as:

4 C₃H₅N₃O₉  → 6 N₂ + 12 CO₂ + 10 H₂O + O₂

From the question, mol ratio of C₃H₅N₃O₉ to O₂ is 4:1. This ratio implies that for every 4 moles of C₃H₅N₃O₉ (nitroglycerine) that decomposes, only 1 mole of O₂ is produced.

Thus:

Moles of O₂ = 1/4 x Moles of C₃H₅N₃O₉ = 1/4 x 1.1013 mol = 0.2753 mol

Mass of O₂ = Moles of O₂ x Molar mass of O₂ = 0.2753 mol x 32 g/mol = 8.81 g

b) The maximum amount of O₂ that can be produced if all the nitroglycerine decomposed completely is known as the theoretical yield.

Actual yield is the real amount of O₂ that was collected when the experiment was carried out.

Given:

Theoretical yield = 8.81 g

Actual yield = 7.84 g

% yield = Actual yield/Theoretical yield x 100 = 7.84 g/8.81 g x 100 = 88.99%