Nitroglycerin detonates according to the balanced reaction:

4 C3H5N3O9 (l) → 12 CO2 (g) + 6 N2 (g) + O2 (g) + 10 H2O (g).

If 150.0 grams of nitroglycerin is detonated, how many L of products (assume all gases at standard temperature and pressure, so 1.00 mole = 22.4 L of volume) are released?

Question

Otto

Chemistry

7 May, 23:55

Nitroglycerin detonates according to the balanced reaction:

4 C3H5N3O9 (l) → 12 CO2 (g) + 6 N2 (g) + O2 (g) + 10 H2O (g).

If 150.0 grams of nitroglycerin is detonated, how many L of products (assume all gases at standard temperature and pressure, so 1.00 mole = 22.4 L of volume) are released?

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Answers (1)

Know the Answer?

Ismael Delgado · Answer 1

From the equation,

4 moles of Nitroglycerin release 29 moles of gasses

Moles of Nitroglycerin present = mass / Mr

150 / (12 x 3 + 5 + 14 x 3 + 16 x 9)

= 0.661

Moles of gasses produced = 29 x 0.661

= 19.2 moles

Liters of gas produced = 19.2 x 22.4

= 430 Liters

Nitroglycerin detonates according to the balanced reaction:4 C3H5N3O9 (l) → 12 CO2 (g) + 6 N2 (g) + O2 (g) + 10 H2O (g).If 150.0 grams of nitroglycerin is detonated, how many L of products (assume all gases at standard temperature and pressure, so 1.00 mole = 22.4 L of volume) are released?

Nitroglycerin detonates according to the balanced reaction:

4 C3H5N3O9 (l) → 12 CO2 (g) + 6 N2 (g) + O2 (g) + 10 H2O (g).

If 150.0 grams of nitroglycerin is detonated, how many L of products (assume all gases at standard temperature and pressure, so 1.00 mole = 22.4 L of volume) are released?