103.5 mL of hydrogen are collected over water at 22.2 degrees Celsius and 758.2 torr from reaction with 0.1356 grams of alloy. calculate the number of moles of dry hydrogen gas that were collected.
Use PV = nRT
P dry H2 = 758.2 torr - x torr water vapor. You will need to look up the vapor pressure of water at 22.2 C. Next, convert pressure in torr to atm by dividing torr by 760.
Don't forget to change V to liters and T to Kelvin. Calculate n, number of moles dry hydrogen.
I got
0.997atm(0.1035L)=n(0.0821)(295.35k)
n=0.0043 mol of H2
How would you calculate N H2, the number of moles of hydrogen gas per gram of alloy?
never mind. I think I got it. Do we just convert that to grams of H2 then divide it by the grams of alloy from the reaction?
JISHKA-BOB
To calculate the number of moles of dry hydrogen gas collected, we need to first determine the partial pressure of hydrogen gas. The given pressure, 758.2 torr, is the total pressure of the gas mixture, which includes the vapor pressure of water.
To find the vapor pressure of water at 22.2 degrees Celsius, we can refer to a vapor pressure table or use the Antoine equation. For simplicity, let's use the vapor pressure of water at 22.2 degrees Celsius, which is approximately 19.8 torr.
Now we can calculate the partial pressure of hydrogen gas:
Partial pressure of hydrogen gas = Total pressure - Vapor pressure of water
= 758.2 torr - 19.8 torr
= 738.4 torr
Next, we need to convert the volume of hydrogen gas collected over water to the volume of dry hydrogen gas. We can use Dalton's Law of Partial Pressure, which states that the total pressure of a gas mixture is the sum of the partial pressures of the individual gases. In this case, the total pressure is the same as the partial pressure of hydrogen gas:
Partial pressure of hydrogen gas = Total pressure of gas mixture
So, the volume of dry hydrogen gas is the same as the volume of hydrogen gas collected over water, which is 103.5 mL.
Now we can calculate the number of moles of dry hydrogen gas using the Ideal Gas Law:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L*atm/(mol*K))
T = temperature (in Kelvin)
Since we have the pressure in torr and volume in mL, we need to convert them to atm and liters respectively.
1 atm = 760 torr (approximately)
1 L = 1000 mL
Converting units:
Partial pressure of hydrogen gas = 738.4 torr * (1 atm / 760 torr)
= 0.97 atm
Volume of dry hydrogen gas = 103.5 mL * (1 L / 1000 mL)
= 0.1035 L
Using the Ideal Gas Law:
0.97 atm * 0.1035 L = n * 0.0821 L*atm/(mol*K) * (22.2 + 273.15) K (converting Celsius to Kelvin)
Simplifying:
0.100,3955 = n * 22.219015
n = 0.004514 moles
Therefore, the number of moles of dry hydrogen gas collected is approximately 0.004514 moles.