Question

Nitroglycerine decomposes violently according to the unbalanced chemical equation below. How many total moles of gases are produced from the decomposition of 1.00 mol C3H5(NO3)3?C3H5(NO3)3 --> CO2(g) + N2(g) + H2O(g) + O2(g)

4.00 mol
6.50 mol
7.25 mol
16.5 mol
29.0 mol

Answer 1

Answer: The total number of moles of gases produced are 7.25 moles.

Step-by-step explanation:

The chemical reaction for the decomposition of
`C_3H_5(NO_3)_3` follows the equation:

`4C_3H_5(NO_3)_3\rightarrow 12CO_2(g)+6N_2(g)+10H_2O(g)+O_2(g)`

We are given:

Moles of
`C_3H_5(NO_3)_3` = 1.00 mol

• For carbon dioxide:

By Stoichiometry of the reaction:

4 moles of
`C_3H_5(NO_3)_3` is producing 12 moles of carbon dioxide.

So, 1 mole of
`C_3H_5(NO_3)_3` will produce =
`(12)/(4)* 1=3mol` of carbon dioxide.

Moles of carbon dioxide produced = 3 moles

• For nitrogen:

By Stoichiometry of the reaction:

4 moles of
`C_3H_5(NO_3)_3` is producing 6 moles of nitrogen.

So, 1 mole of
`C_3H_5(NO_3)_3` will produce =
`(6)/(4)* 1=1.5mol` of nitrogen.

Moles of nitrogen produced = 1.5 moles

• For water:

By Stoichiometry of the reaction:

4 moles of
`C_3H_5(NO_3)_3` is producing 10 moles of water.

So, 1 mole of
`C_3H_5(NO_3)_3` will produce =
`(12)/(10)* 1=2.5mol` of water.

Moles of water produced = 2.5 moles

• For oxygen:

By Stoichiometry of the reaction:

4 moles of
`C_3H_5(NO_3)_3` is producing 1 moles of oxygen.

So, 1 mole of
`C_3H_5(NO_3)_3` will produce =
`(1)/(4)* 1=0.25mol` of oxygen.

Moles of oxygen produced = 0.25 moles

Total number of moles of the gases produced are = 3 + 1.5 + 2.5 + 0.25 = 7.25 moles.

Hence, the total number of moles of gases produced are 7.25 moles.