Question
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.
Answers
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.
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?
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
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