90 nitroglycerin (c3h5n3o9) is a powerful explosive. its decomposition may be represented by this reaction generates a large amount of heat and many 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 2.00 × 102 g of nitroglycerin? (b) calculate the percent yield in this reaction if the amount of o2 generated is found to be 6.55 g.

The reaction for the decomposition of nitroglycerin is:

4 C₃H₅N₃O₉ (l) → 6 N₂ (g) + 12 CO₂ (g) + 10 H₂O (g) + O₂ (g)

a.) The molar mass of C₃H₅N₃O₉ is 227 g/mol. Let's compute for the number of moles nitroglycerin.

Moles of nitroglycerin = 200 g * 1 mol/227 g = 0.881 mol
Moles O₂ produced = 0.881 mol nitroglycerin(1 mol O₂/4 mol nitroglycerin) = 0.22 mol O₂

The molar mass of O₂ is 32 g/mol. So, the corresponding mass would be:
Mass O₂ produced = 0.22 mol O₂(32 g/mol) = 7.048 g O₂

b.) The percent yield is equal to the actual yield divided by the theoretical yield. The amount calculated in part a is the theoretical yield, while the given in part b is the actual yield.

Percent yield = (6.55 g/7.048 g)*100 = 92.93%

Answer is to avoid experimental error