posted by Marie on .
How would I go about doing this problem?
146 mL of wet H2 gas is collected over water at 29.0 degrees C and room pressure of 750 torr. Give the volume of the dry H2 gas collected at STP, give the mass of the dry H2 collected, and give the volume of the dry H2 at room conditions and the volume of H2O in the collected gas at room conditions.
I can get you started.
(P1V1/T1) = (P2V2/T2)
Plug in current and STP conditions to correct H2 to STP conditions. Remember Ptotal = PH2 + PH2O. Look up the vapor pressure at the conditions under which H2 gas is collected and subtract from Ptotal (750 torr). This will give you the volume of dry H2 gas at STP.
You can use the PV/T equation again but substitute the room conditions for that part of the problem.
From the volume at STP you know that 22.4 L is 1 mol; therefore, volume in L/22.4 = # moles H2 gas and moles x molar mass = grams.
Check my thinking. This is a long problem.