Question

A gaseous mixture consists of 28.4 mole percent of hydrogen and 71.6 mole percent of methane. A 15.6-L gas sample, measured at 19.4°C and 2.23 atm. is burned in air. Calculate the heat released.

Answer 1

Answer on Question #76749, Chemistry / General Chemistry

Question:

A gaseous mixture consists of 28.4 mole percent of hydrogen and 71.6 mole percent of methane. A 15.6 L gas sample, measured at $19.4^{\circ}\mathrm{C}$ and 2.23 atm is burned in air. Calculate the heat released.

Solution:

Pressure: $p = 2.23 \cdot 101325 = 225955 \, \text{Pa}$

Volume: $V = 15.6 \, \text{L} = 0.0156 \, \text{m}^3$

Temperature: $T = 19.4 + 273.1 = 292.5 \, \text{K}$

Gas constant: $R = 8.314 \, (\text{m}^3 \cdot \text{Pa}) / (\text{mol} \cdot \text{K})$

Ideal gas low: $pV = nRT$, so the total amount of molecules of gases:

$$n = pV / RT = (225955 \cdot 0.0156) / (8.314 \cdot 292.5) = 1.45 \, \text{mol}$$

Amount of hydrogen: $1.45 \cdot 0.284 = 0.4118 \, \text{mol}$

Heat of combustion of hydrogen: $286 \, \text{kJ/mol}$

Energy released by hydrogen: $286 \cdot 0.4118 = 117.77 \, \text{kJ}$

Amount of methane: $1.45 \cdot 0.716 = 1.0382 \, \text{mol}$

Heat of combustion of methane: $889 \, \text{kJ/mol}$

Energy released by methane: $889 \cdot 1.0382 = 922.96 \, \text{kJ}$

Total heat released: $117.77 + 922.96 = 1040.73 \, \text{kJ}$

Answer:

1040.73 kJ