Question

[COLLECTING GASES OVER WATER]

2 H2O(L) --> 2 H2(g) + O2(g)

Imagine a gas collection apparatus immersed in a water bath with an electrode inside of it that sticks into the water, so all H2(g) and O2(g) produced is collected. After performing the reaction, you find that you've collected 625 mL of gas and you measure the temperature to be 40degC. If the pressure inside the collection apparatus is 1130 mmHg, how many mL of H2O(L) was electrolyzed in the reaction? The vapor pressure of water at 40degC is 55.3 mmHg. (Assume the density of water is 1 g/mL.)

Answer 1

I've tried numerous times and keep getting that the answer is above 600 mL of water, but the answer the professor gives is about 0.414 mL H2O. For some reason, I have not had to factor in the density of water, which gives me the feeling that I'm doing something wrong. Can someone help?

Answer 2

whenever you collect a gas over water, a small amount of water vapor is collected along with the gas of interest. If you collect enough "wet" gas so that the total pressure is 1130mmHg, a \(small \space portion\) of it will be water vapor. The vapor pressure of water tells you how much.

You need to subtract the vapor pressure of water from the total pressure of 1130mmHg to find the pressure of the "dry" gas

Answer 3

then the ratio of the pressures will equal the ratio of the volumes