Nitroglycerin,used both in medicine and as an explosive,can be prepared by the carefully controlled reaction of glycerol (C₃H₈O₃) with nitric acid,as symbolized by:
C₃H₈O₃ + 3 HNO₃ → C₃H₅N₃O₉ + 3 H₂O
What mass of nitric acid is required for the production of 2.8 g of nitroglycerin by a process having a 67% yield?

Question

Quizplus · Accepted Answer

First, calculate the theoretical yield of nitroglycerin from 2.8 g based on a 67% actual yield: $2.8 \, \text{g} / 0.67 = 4.18 \, \text{g}$ theoretical yield. The molar mass of nitroglycerin (C3H5N3O9) is approximately 227 g/mol, and that of nitric acid (HNO3) is about 63 g/mol. From the balanced equation, 3 moles of HNO3 are required to produce 1 mole of nitroglycerin. Therefore, the mass of HNO3 needed for the theoretical yield of 4.18 g of nitroglycerin is calculated as follows: $(4.18 \, \text{g} / 227 \, \text{g/mol}) \times 3 \times 63 \, \text{g/mol} = 3.5 \, \text{g}$.

Nitroglycerin,used Both in Medicine and as an Explosive,can Be Prepared