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.

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 = $\frac{12}{4}\times 1=3mol$ 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 = $\frac{6}{4}\times 1=1.5mol$ 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 = $\frac{12}{10}\times 1=2.5mol$ 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 = $\frac{1}{4}\times 1=0.25mol$ 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.