# A 60.0 mL sample of #CO_2# gas is collected over water at 70.0° and 101.3 kPa. What is the volume of the dry gas at STP?

##### 1 Answer

#### Explanation:

The important thing to remember about gases that are being collected **over water** is that the total pressure measured **includes** the vapor pressure of water at that given temperature.

In this case, the collection tube will hold a **gaseous mixture** that contains molecules of carbon dioxide,

This means that you can use **Dalton's Law of Partial Pressures** to determine the partial pressure of carbon dioxide in this mixture.

#color(blue)(P_"total" = sum_i P_i)" "# , where

In this case, the vapor pressure of water at

http://www.endmemo.com/chem/vaporpressurewater.php

This means that you have

#P_"total" = P_(CO_2) + P_(H_2O)#

#P_(CO_2) = "101.3 kPa" - "31.09 kPa" = "70.21 kPa"#

Now, **STP** conditions imply a temperature of *volume* of the gas at STP, and keeping in mind that the number of moles of gas remains constant, you can use the combined gas law equation

#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "# , where

Plug in your values and solve for **do not** forget to convert the temperature from *degrees Celsius* to *Kelvin*!

#V_2 = P_1/P_2 * T_2/T_1 * V_1#

#V_2 = (70.21 color(red)(cancel(color(black)("kPa"))))/(100.0color(red)(cancel(color(black)("kPa")))) * ( (273.15 + 0)color(red)(cancel(color(black)("K"))))/((273.15 + 70.0)color(red)(cancel(color(black)("K")))) * "60.0 mL"#

#V_2 = color(green)("33.5 mL")#

The answer is rounded to three sig figs.