A mixture in which the mole ratio of hydrogen to oxygen is (exactly) 2:1 is used to prepare water by the reaction 2 H2 (g) + O2 (g) → 2 H2O (g) The total pressure in the container is 0.951 atm at 22°C before the reaction. What is the final pressure in the container after the reaction, with a final temperature of 125°C, no volume change, and an 85.0% yield?

Explanation:

Hello,

In this case, it is convenient to state the ideal gas law in terms of the undergoing change due to the chemical reaction:

Since no volume change is present, one obtains:

Thus, it is possible to assume that the moles of hydrogen and oxygen are 2 mole and 1 mole respectively, in order to respect the given ratio, thus, the yielded moles of water turns out:

It is important to notice that just the 85.0% of the theoretical moles are actually produced, thus:

In such a way, there is a remaining amount of hydrogen and oxygen that are computed via:

Thus, since the initial volume is:

And it is conserved, at the new temperature, the pressure is:

Best regards.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2