4C₃H₅(NO₃)₃[tex] _{(l)} [/tex] ------> 12CO₂[tex]_{(g)} [/tex] + 6N₂[tex]_{(g)} [/tex] + 10H₂O[tex]_{(g)} [/tex] + O₂[tex]_{(g)} [/tex]
mol of CO₂ = [tex] \frac{mass}{molar mass} [/tex]
= [tex] \frac{25g}{44.01 g/mol} [/tex]
mol ratio of CO₂ : C₃H₅(NO₃)₃
12 : 4
∴ if mole of CO₂ = 0.568 mol
then " " C₃H₅(NO₃)₃ = [tex] \frac{0.568 mol}{12} * 4[/tex]
= 0.189 mol
∴ mass of nitroglycerin = mole * Mr
= 0.189 mol * 227.0995 g / mol
= 43.00 g
First, we have to calculate the moles of
Moles of CO2 = mass of CO2/Molar mass of CO2 = 25g/44.01g/mole =0.57 moles
Now we have to calculate the moles of nitroglycerin.
The balanced chemical reaction is,
4C3H5(NO3)3(l)~ 12CO2(G) + 6N2(g) +)2(g)
From the balanced reaction, we conclude that
As 12 moles of CO2 were obtained from 4 moles C3H5(NO3)3 of
So, 0.57 moles of CO2 were obtained from 4/12 x 0.57 = 0.19 the mole of
C3H5(NO3)3
Now we have to calculate the mass of nitroglycerin.
Mass ofC3H5 (NO3)3 = Moles of C3H5(NO3)3 x Molar masse of C3H5(NO3)3
Mass ofC3H5 (NO3)3 = (0.19mole) x (227.0995g/mole) = 43.15g
Therefore, the mass of nitroglycerin will be, 43.15 grams.
Nitroglycerin is extremely sensitive to shock and to rapid heating; it begins to decompose at 50–60 °C (122–140 °F) and explodes at 218 °C (424 °F). The safe use of nitroglycerin as a blasting explosive became possible after the Swedish chemist Alfred B.
Learn more about nitroglycerin at
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