A vessel contains 16 g of hydrogen and 128 g of oxygen at standard temperature and pressure. The volume of the vessel in cm$$^3$$ is:

A vessel contains 16 g of hydrogen and 128 g of oxygen at STP.

Moles of H$$_2$$ = $$\frac{16}{2} = 8$$ mol

Moles of O$$_2$$ = $$\frac{128}{32} = 4$$ mol

Total moles = $$8 + 4 = 12$$ mol

At STP (Standard Temperature and Pressure: 273.15 K, 1 atm), 1 mole of ideal gas occupies 22,400 cm$$^3$$.

$$V = 12 \times 22400 = 268800 \text{ cm}^3 = 2.688 \times 10^5 \text{ cm}^3$$

This is approximately $$27 \times 10^4$$ cm$$^3$$.

Hence, the correct answer is Option C: $$27 \times 10^4$$.