°C and 1 atm pressure. The resulting mixture is ignited and allowed to react completely. What is the final pressure in the container at the same temperature?

We can start by writing the balanced chemical equation for the combustion of ethane:

C2H6 + 3O2 → 2CO2 + 3H2O

This equation tells us that one molecule of ethane reacts with three molecules of oxygen to produce two molecules of carbon dioxide and three molecules of water. However, we don't have any oxygen in our initial mixture - we only have ethane and hydrogen.

Fortunately, we can use the ideal gas law to help us solve this problem. The ideal gas law relates the pressure, volume, temperature, and number of moles of a gas:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Since we have equal weights of ethane and hydrogen, we can assume that we have equal numbers of moles of each gas. Let's call this number "n". We can also assume that the volume of the container is constant throughout the reaction, so we can write:

P_initialV = 2nRT_initial

where P_initial is the initial pressure and T_initial is the initial temperature.

After the reaction, we have two products: carbon dioxide and water. These are both gases, so we can assume that they mix together to form a single gas that occupies the same volume as the initial mixture. This means that the final number of moles in the container is equal to the sum of the moles of carbon dioxide and water produced by the reaction:

n_final = 2n + 3n = 5n

To find the final pressure, we can use the ideal gas law again:

P_finalV = 5nRT_final

where P_final is the final pressure and T_final is the final temperature (which is the same as the initial temperature, since the reaction takes place at constant temperature).

We can solve for P_final by dividing both sides of the equation by V and substituting the expression we found earlier for V:

P_final = 5nRT_final / (P_initialV)

We can simplify this expression by using the fact that the initial pressure is 1 atm, and that one mole of any gas at 25°C and 1 atm occupies a volume of about 24.5 L (this is known as the molar volume of an ideal gas at standard temperature and pressure):

P_final = 5nRT_final / (24.5 L)

We can also substitute the values of R and T in SI units:

P_final = 5n(8.31 J/(mol*K))(298 K) / (24.5 L)

P_final = 5n(2.46 J/L) = 12.3n J/L

Finally, we can substitute n = m/M, where m is the mass of each gas (which is equal since we have equal weights) and M is the molar mass of each gas:

P_final = 12.3(m/M) J/L

For ethane, M = 30 g/mol, and for hydrogen, M = 2 g/mol. Therefore, the final pressure is:

P_final = 12.3(1/16 + 1/30) J/L

P_final =