So the question I have is:
Calculate the pressure of dry hydrogen collected. (Use the water temperature).
And the evidence that we obtained was:
Length of magnesium
3.8Cm
Mass of 1 m of magnesium
1.254g
Volume of hydrogen collected
67mL
Air temperature
27C
Water temperature
26C
Air pressure
95.97kPa
Vapour pressure of water
3.36kPa
I really don't understand it, thank you!
I thought that would be the next question. :-)
Use P1V1 = P2V2. That will give you the volume at STP if you plug in values for STP of 273 K and 101.325 kPa.
Remember Dalton's Law of Partial Pressures; i.e., the total pressure of a system is the sum of the partial pressures of each gas. So
Total P = pH2O vapor + pH2(dry) gas.
You know total pressure--that's air pressure of 95.97 kPa. You are given the pH2O at that temperature of 3.36 kPa. That lets you calculate the pH2. That will be the pressure of dry H2O since you've subtracted the pressure of the water vapor.
Thank you! (: I'm also confused about the 2nd question:
Correct your dry hydrogen volume to a volume at STP (Assume the temperature of the hydrogen in your collection tube was the same as the air temperature).
Sorry for all the questions. So the equation I ended up with was:
101.325 x 0.067 / 95.97 = V2
But I'm assuming that's not right, so instead of x0.067 would it be x 273K?
No but that's my mistake. I didn't write all of the equation.
It should be (P1V1/T1) - P2V2/T2)
Substitute and solve for V2.
To calculate the pressure of dry hydrogen collected, you can use the ideal gas law equation, which is:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant
T = temperature of the gas in Kelvin
In this case, we need to calculate the pressure of dry hydrogen. Dry hydrogen means that there is no water vapor present in the collected gas.
To begin the calculation, let's convert the given temperatures from Celsius to Kelvin:
Air temperature = 27°C + 273.15 = 300.15 K
Water temperature = 26°C + 273.15 = 299.15 K
Next, let's convert the volume of hydrogen collected from milliliters to liters, since the ideal gas law requires volume to be in liters:
Volume of hydrogen collected = 67 mL = 67/1000 L = 0.067 L
Now, we need to determine the number of moles of hydrogen. To do this, we will use the relationship between the mass of magnesium and the moles of hydrogen produced.
First, we need to calculate the molar mass of magnesium (Mg). The molar mass of Mg is 24.31 g/mol.
Next, let's convert the length of magnesium from cm to meters:
Length of magnesium = 3.8 cm = 3.8/100 m = 0.038 m
Then, calculate the mass of magnesium used in the reaction:
Mass of magnesium = mass of 1 m of magnesium = 1.254 g
Now, let's calculate the moles of hydrogen produced using the stoichiometry of the reaction between magnesium and hydrogen:
1 mol of Mg produces 1 mol of H2
So, the moles of hydrogen produced can be calculated as follows:
moles of hydrogen = (mass of magnesium / molar mass of magnesium) * 1
Now, we have all the necessary values to calculate the pressure of dry hydrogen collected.
First, let's substitute the known values into the ideal gas law equation:
PV = nRT
Since we want to find the pressure (P), we rearrange the equation:
P = (nRT) / V
Now, substitute the values into the equation:
P = [(moles of hydrogen * R * T) / V]
The ideal gas constant (R) is 8.314 J/(mol·K). The temperature (T) needs to be in Kelvin, so we'll use the given air temperature (300.15 K).
Finally, substitute the values and calculate the pressure of dry hydrogen collected.
P = [(moles of hydrogen * 8.314 J/(mol·K) * 300.15 K) / 0.067 L]