Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition may be represented by$$4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9} \longrightarrow 6 \mathrm{~N}_{2}+12 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$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 \(\mathrm{O}_{2}\) in grams that can be obtained from \(2.00 \times 10^{2} \mathrm{~g}\) of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of \(\mathrm{O}_{2}\) generated is found to be \(6.55 \mathrm{~g}\).

Step by step solution

01

Compute the molar mass of nitroglycerin (C3H5N3O9)

The molar mass is calculated by adding up the molar masses of all the atoms in one molecule of nitroglycerin. Refer to the periodic table for the molar masses: Carbon (C) is 12.01 g/mol, Hydrogen (H) is 1.01 g/mol, Nitrogen (N) is 14.01 g/mol and Oxygen (O) is 16.00 g/mol. So, \(Molar Mass_{C3H5N3O9} = 3*12.01 + 5*1.01 + 3*14.01 + 9*16.00 = 227.09 g/mol\)

02

Calculate moles of nitroglycerin

Now, to calculate the amount of nitroglycerin in moles supplied in the reaction, use the formula \(moles = mass/molar mass\). So, \(Moles_{C3H5N3O9} = (2.00x10^2 g) / 227.09 g/mol = 0.88 mol\)

03

Calculate the theoretical yield of O2

Using the stoichiometric relationship, for every 4 moles of nitroglycerin, 1 mole of \(O_2\) is produced. So, for 0.88 mol of nitroglycerin, number of moles of \(O_2\) produced can be calculated: \(Moles_{theoretical-yield-O2} = (0.88 mol x 1) / 4 = 0.22 mol\). This is converted into grams using the molar mass of \(O_2 = 32 g/mol\). So, \(Mass_{theoretical-yield-O2} = 0.22 mol x 32 g/mol = 7.04 g\)

04

Calculate percent yield

Now, to calculate the percent yield of the reaction, use the formula \((Actual output / Theoretical output) x 100\). It was given that the actual oxygen yield was 6.55g. So, \(Percent Yield = (6.55 g / 7.04 g) x 100 = 93.03\%\)

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!