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.
To solve this problem, we need to use the ideal gas law and the concept of water vapor pressure. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Step 1: Convert the temperature to Kelvin
Given: T = 29.0 degrees C
To convert to Kelvin, we add 273.15 to the temperature:
T = 29.0 + 273.15 = 302.15 K
Step 2: Calculate the number of moles of wet H2 gas collected
Given: V = 146 mL = 0.146 L
Using the ideal gas law equation, we can solve for the number of moles (n):
PV = nRT
n = PV / RT
n = (750 torr * 0.146 L) / (0.0821 L·atm/mol·K * 302.15 K)
n = 0.00585 mol
Step 3: Calculate the volume of the dry H2 gas at STP
At STP (Standard Temperature and Pressure), the conditions are 273.15 K and 1 atm.
Using the ideal gas law equation, we can calculate the volume (V) of the dry H2 gas:
PV = nRT
V = nRT / P
V = (0.00585 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm
V = 0.132 L
Step 4: Calculate the mass of the dry H2 gas collected
To calculate the mass, we need to know the molar mass of H2, which is 2 g/mol.
m = n * molar mass
m = 0.00585 mol * 2 g/mol
m = 0.0117 g
Step 5: Calculate the volume of the dry H2 gas at room conditions
Given: T = 29.0 degrees C = 302.15 K
Using the ideal gas law equation, we can calculate the volume (V) of the dry H2 gas at room conditions:
PV = nRT
V = nRT / P
V = (0.00585 mol * 0.0821 L·atm/mol·K * 302.15 K) / 750 torr
V = 0.0187 L
Step 6: Calculate the volume of water vapor in the collected gas at room conditions
To calculate the volume of water vapor, we need to subtract the volume of dry H2 gas from the initial volume of wet gas:
Volume of H2O = Initial volume - Volume of dry H2
Volume of H2O = 0.146 L - 0.0187 L
Volume of H2O = 0.1273 L
Therefore, the volume of the dry H2 gas collected at STP is 0.132 L, the mass of the dry H2 collected is 0.0117 g, the volume of the dry H2 gas at room conditions is 0.0187 L, and the volume of H2O in the collected gas at room conditions is 0.1273 L.