Question

For the combustion of hydrazine rocket fuel, N2H4(l), predict the volume of nitrogen dioxide produced at 850°C and 180 kPa, when 50.0 L of oxygen gas (measured at STP) is consumed.

Answer 1

The balanced equation of the chemical reaction between hydrazine N₂H₄ and oxygen O₂ is:

N₂H₄ + 3O₂ $\rightarrow$ 2NO₂ + 2H₂O.

According to the stoichiometry coefficients, when 3 mol of oxygen reacts, 2 mol of nitrogen dioxide is produced:

$\frac{n(O_2)}{3} = \frac{n(NO_2)}{2}$ .

Making use of the ideal gas law ($pV = nRT$) and the definition of the STP conditions (10⁵ Pa, 273.15 K, gas constant 8.314 J mol^-1 K^-1), the number of the moles of the oxygen consumed is:

$n_{O_2} = \frac{pV_{O_2}}{RT} = \frac{10^5(Pa)·50·10^{-3}(m^3)}{8.314(J/(mol K))·273.15(K)}$

$n_{O_2} = 2.2$ mol.

Therefore, the volume of the nitrogen dioxide at 850+273.15=1123.15 K and 180·10³ Pa would be:

$V_{NO_2} = \frac{2}{3}·n_{O_2} \frac{RT}{p}$

$V_{NO_2}= \frac{2}{3}·2.2· \frac{8.314 (J/(molK))·1123.15(K)}{180·10^3(Pa)} = 76·10^{-3}$ m³, or 76 L.

Answer: 76 L of nitrogen dioxide is produced.