Sometimes in lab we collect the gas formed by a chemical reaction over water (see sketch at right). This makes it easy to isolate and measure the amount of gas produced. Wala Suppose the H, gas evolved by a certain chemical reaction taking place at 40.0 °C is collected over water, using an apparatus something like that in the sketch, and the final volume of gas in the collection tube is measured to be 80.0 mL. Sketch of a gas-collection apparatus Calculate the mass of H, that is in the collection tube. Round your answer to 2 significant digits. You can make any normal and reasonable assumption about the reaction conditions and the nature of the gases. XS?

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Explanation:

Case 1) When we consider vapor pressure of H2O;

Let the pressure is 1 atm OR 760 torr.

As H2 is collected over water, we have to consider the vapor pressure of H2O as well.

Using data i.e. vapor pressure of H2O at 40° C = 55.365 torr

So, pressure of H2 = P = 760 - 55.365 = 704.635 torr = 704.635/760 = 0.9272 atm

Volume of H2 = 80 ml = 0.08 liter

Temperature (T) = 40 + 273 = 313 K

Gas constant (R) = 0.0821 L atm/mol K

Let n is moles of H2. Applying ideal gas equation;

PV = nRT

n = PV/RT = 0.9272 * 0.08 / 0.0821 * 313

n = 0.00289 moles

Mass of H2 = moles * molar mass = 0.00289 * 2.016 = 0.00582 grams

OR 5.8*10^-3 grams ...Answer

> Case 2) When we don't consider vapor pressure;

Pressure of H2 = 1 atm, all other parameters will remain same as in case 1.

So, mass of H2 = 6.3*10^-3 grams

But Case 1) is correct approach as in question it is mentioned that H2 is collected over water.

i do not get it   at all