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?

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

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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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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?