Answer to Question #122381 in Chemistry for aljoc

Question

The reaction of 2,4,6-trinitrotoluene (TNT) and oxygen is given by the equation: 4 C7H5(NO2)3(l) + 21O2(g) → 28 CO2(g) + 6 N2(g) + 10 H2O(g)
Assuming all gases are measured at SATP, what volume of carbon dioxide is formed when 10.0 L of oxygen gas is consumed?

Answer 1

The balanced reaction equation of 2,4,6-trinitrotoluene (TNT) and oxygen is:

4 C₇H₅(NO₂)₃(l) + 21O₂(g) → 28 CO₂(g) + 6 N₂(g) + 10 H₂O(g).

As one can see, when 21 mol of oxygen is consumed, 28 mol of carbon dioxide is formed. Therefore, the number of the moles of oxygen reacted and carbon dioxide formed relate as:

$21·n(CO_2) = 28·n(O_2)$ .

According to the ideal gas law ($pV = nRT$), at SATP conditions:

$\frac{n(CO_2)}{V(CO_2)} = \frac{n(O_2)}{V(O_2)}$ .

Making use of the number of the moles relation:

$\frac{28}{21} = \frac{V(CO_2)}{V(O_2)}$

$V(CO_2) = 10(L)·\frac{28}{21} = 13.3$ L.

Answer: 13.3 L of carbon dioxide is formed when 10.0 L of oxygen gas is consumed.