Question

Suppose that 10.0 L of Carbon Dioxide gas are produced by this reaction, 4C3H5N3O9 -> 12 CO2 + 10H2O + 6N2 +O2, at a temperature of -5 degrees C, and a pressure of exactly 1 atm. Calculate the mass of nitroglycerin that must have reacted in grams.

Answer 1

The mass of nitroglycerin : 34.52 g

Further explanation

Reaction

4C₃H₅N₃O₉ ⇒ 12 CO₂ + 10H₂O + 6N₂ +O₂

Volume = 10 L

Temperature = -5°C=268 °K

Pressure = 1 atm

mol of CO₂ (ideal gas) :

$\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 10}{0.082\times 268}\\\\n=0.455$

mol ratio C₃H₅N₃O₉ : mol CO₂= 4 : 12, so mol C₃H₅N₃O₉ :

$\tt \dfrac{4}{12}\times 0.455=0.152$

mass C₃H₅N₃O₉ (MW=227,0865 g/mol):

$\tt 0.152\times 227.0865=\boxed{\bold{34.52~g}}$