if 22.4 g of TNT is combusted in a 0.500 L container filled with O2 at a pressure of 8.58 bar and a temperature of 298 K, calculate the maximum mass of CO2 that could be produced.

10.12 g of CO2

Explanation:

The equation of the reaction is;

2C7H5N3O6(s) + 21/2O2(g) → 14CO2 (g) + 5H2O(g) + 3N2(g)

From the information provided;

P= 8.58 bar = 8.5 atm

V =0.5 L

T = 298 K

R = 0.082 atmLK-1Mol-1

n= ?

From

PV =nRT

n = PV/RT

n= 8.5 * 0.5/0.082 * 298

n= 4.25/24.44

n= 0.17 moles of oxygen

Molar mass of TNT =227.13g/mol

The number of moles of TNT reacted = 22.4g/227.13g/mol= 0.099 moles of TNT

We determine the limiting reactant. This is the reactant that yields the least amount of CO2

From the reaction equation;

10.5 moles of oxygen yields 14 moles of CO2

0.17 moles of oxygen yields 0.17 * 14/10.5 = 0.23 moles of CO2

For TNT;

2 moles of TNT yields 14 moles of CO2

0.099 moles of TNT yields 0.099 * 14/2 = 0.693 moles of CO2

Oxygen is the limiting reactant

Therefore

Maximum mass of CO2 = 0.23 moles of CO2 * 44 g/mol = 10.12 g of CO2

a c d and f

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