The pressure exerted by a non-reactive gaseous mixture of 6.4 g of methane and 8.8 g of carbon dioxide in a 10 L vessel at 27°C is ________ kPa. (Round off to the Nearest Integer).
[Assume gases are ideal, R = 8.314 J mol$$^{-1}$$ K$$^{-1}$$, Atomic masses: C: 12.0 u, H: 1.0 u, O: 16.0 u]

First, calculate the moles of each gas. The molar mass of methane ($$CH_4$$) is $$12 + 4 \times 1 = 16$$ g/mol, so moles of methane = $$\frac{6.4}{16} = 0.4$$ mol. The molar mass of carbon dioxide ($$CO_2$$) is $$12 + 2 \times 16 = 44$$ g/mol, so moles of carbon dioxide = $$\frac{8.8}{44} = 0.2$$ mol.

The total moles of gas in the mixture is $$n_{total} = 0.4 + 0.2 = 0.6$$ mol. The temperature in Kelvin is $$27 + 273 = 300$$ K.

Using the ideal gas law $$PV = nRT$$, the total pressure is $$P = \frac{nRT}{V} = \frac{0.6 \times 8.314 \times 300}{10} = \frac{1496.52}{10} = 149.652$$ kPa. Rounding to the nearest integer gives $$P \approx 150$$ kPa.