My question concerns the stoichiometry of gases.

Question

Nitroglycerine, C3H5N3O9, is an oily liquid at room temperature that explodes just by being jostled too much. When it explodes, the following reaction occurs: 4C3H5N3O9(l) yields 10H2O(g) + 12CO2(g) + 6N2(g) + O2(g). Assume that in such an explosion 1.00 g H2O is produced at 350 degrees C and 50.0 bar. Determine:

a. the total volume of all the reaction products. 

b. the mass of nitroglycerine consumed in the reaction. 

Answer 1

4C₃H₅N₃O₉(l) ==> 10H₂O(g) + 12CO₂(g) + 6N₂(g) + O₂(g)

Given: 1.00 g H₂O produced @ 350ºC and 50.0 bar of pressure

Determine: total volume of all products (gases)

Procedure:

Convert 1.00 g H₂O to moles

Convert 350ºC to Kelvin = 350 + 273 = 623K

Convert 50.0 bar to atmospheres = 50.0 bar x 1 atm / 1.01325 bar = 49.35 atm

Use ideal gas law to find volume of H₂O(g)

Then use stoichiometry to find amount of all products and the mass of nitroglycerin

1.00 g H₂O x 1 mol H₂O / 18.0 g = 0.05556 mols H₂O

PV = nRT .. ideal gas law

V = volume of H₂O(g) = nRT/P = (0.05556 mol)(0.0821 Latm/Kmol)(623K) / 49.35 atm

V = 0.0576 Liters of H₂O formed

Volume CO₂(g) = 0.0576 L H₂O x 12 L CO₂ / 10 L H₂O = 0.0691 L CO₂

Volume N₂(g) = 0.0576 L H₂O x 6 L N₂ / 10 L H₂O = 0.0346 L N₂

Volume O₂(g) = 0.0576 L H₂O x 1 L O₂ / 10 L H₂O = 0.00576 L O₂

Total volume of all products = 0.0576 L + 0.0691 L + 0.0346 L + 0.00576 L = 0.167 Liters

moles of C₃H₅N₃O₉ used = 0.05556 mols H2O x 4 mol C₃H₅N₃O₉ / 10 mols H2O = 0.0222 mol C₃H₅N₃O₉

mass C₃H₅N₃O₉ = 0.0222 mols C₃H₅N₃O₉ x 227 g / mol = 5.04 g C₃H₅N₃O₉ consumed