Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition can 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

Calculate the Molar Mass of Nitroglycerin and Oxygen

First, one must calculate the molar mass of Nitroglycerin \((C_{3}H_{5}N_{3}O_{9})\) and of \(O_{2}\). The molar mass of Nitroglycerin can be calculated as: \[227.09 \, \text{g/mol} = 3(12.01 \, g/mol) + 5(1.01 \, g/mol) + 3(14.01 \, g/mol) + 9(16.00 \, g/mol)\]The molar mass of \(O_{2}\) is:\[32.00 \, \text{g/mol}\]

02

Calculate the maximum amount of \(O_{2}\) produced

Using stoichiometry, we have that 4 moles of Nitroglycerin produce 1 mole of \(O_{2}\). Thus, we can calculate the number of moles of Nitroglycerin in 200g:\[0.88 \, \text{mol} = 200 \, \text{g}/227.09 \, \text{g/mol}\]By stoichiometry, this would yield:\[0.22 \, \text{mol} = 0.88 \, \text{mol}/4\]Converting this many moles of \(O_{2}\) to grams (given its molar mass of 32.00 g/mol) we obtain:\[7.04 \, \text{g} = 0.22 \, \text{mol} \times 32.00 \, \text{g/mol}\]

03

Calculating the percent yield

The percent yield can be calculated as:\[\text{Percent Yield} = (\text{Actual Yield}/\text{Theoretical Yield}) \times 100\%\]Given the problem, the Actual Yield is 6.55g while the Theoretical Yield is 7.04g. Replacing these into the formula yields:\[\text{Percent Yield} = (6.55 g/7.04 g) \times 100\% = 93.03\%\]

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