Question

Assume that you have 1.15 g of nitroglycerin in a 579.5 mL steel container at 20.0 ∘C and 1.00 atm pressure. An explosion occurs, raising the temperature of the container and its contents to 425 ∘C. The balanced equation is
4C3H5N3O9(l)→12CO2(g)+10H2O(g)+6N2(g)+O2(g)
(Figure 1)

Part A

How many moles of nitroglycerin were in the container originally?

Part B

How many moles of gas (air) were in the container originally?

Part C

How many moles of gas are in the container after the explosion?

Part D

What is the pressure (in atmospheres) inside the container after the explosion according to the ideal gas law?

Answer 1

Part a

Mass of nitroglycerin = 1.15 g

Molecular weight of nitroglycerin (C3H5N3O9)

= 227.08 g/mol

Moles of nitroglycerin = mass/molecular weight

= (1.15 g) / (227.08 g/mol)

= 0.00506 mol

= 5.06 x 10^-3 mol

Part b

From the ideal gas equation

Pressure x volume = moles x gas constant x temperature

1 atm x 579.5 mL x 1L/1000 mL = n x 0.0821 L-atm/mol-K x (273 + 20)K

Moles of air n = 0.0241 mol

Part c

After explosion

From the stoichiometry of the reaction

Moles of reactants (C3H5N3O9) = 4 mol

Moles of products = 12 + 10 + 6 + 1 = 29 mol

Actual Moles of nitroglycerin = 5.06 x 10^-3 mol

Moles of gaseous products

= (5.06 x 10^-3 mol) x (29 mol gas) / (4 mol gas)

= 0.036685 mol

Total moles of products = moles of gas + moles of air

= 0.036685 + 0.0241

= 0.060785 mol

Part d

After explosion

Temperature T = 425 + 273 = 698 K

From the ideal gas equation

PV = nRT

P = (0.060785 mol) x (0.0821 L-atm/mol-K) x (698K)/(579.5 mL x 1L/1000 mL)

P = 6.01 atm