Question #aae8e
1 Answer
Explanation:
The idea here is that the gas produced by a chemical reaction is often collected over water, which implies that it is mixed with water vapor.
In order to find the partial pressure of the dry gas, which is the gas produced by the reaction without the presence of the water vapor, you must use the partial pressure of water vapor at the temperature at which the gas is being collected.
In other words, you must use Dalton's law of Partial Pressures, which states that the partial pressure of a gaseous mixture is equal to the sum of the partial pressure of its gaseous components.
In your case, you will have
P_"mixture" = P_"dry gas" + P_"water vapor"Pmixture=Pdry gas+Pwater vapor
which will get you
P_"dry gas" = P_"mixture" - P_"water vapor"Pdry gas=Pmixture−Pwater vapor
Now, convert the temperature to degrees Celsius
t[""^@"C"] = "308 K" - "273.15 K" = 34.85^@"C"t[∘C]=308 K−273.15 K=34.85∘C
Now look up the partial pressure of water vapor at this temperature
http://www.endmemo.com/chem/vaporpressurewater.php
You'll find it listed as
P_"water vapor" = "41.72 mmHg"Pwater vapor=41.72 mmHg
which is equivalent to
P_"water vapor" = "41.72 torr"Pwater vapor=41.72 torr
You can now say that the partial pressure of the dry gas is equal to
P_"dry gas" = "742 torr" - "41.72 torr"Pdry gas=742 torr−41.72 torr
color(darkgreen)(ul(color(black)(P_"dry gas" = "700. torr")))
The answer is rounded to three sig figs.
