Hydrogen gas produced by this reaction is typically collected via water displacement, during which time the hydrogen gas becomes saturated with water vapor. If 238.4 mL of gas with a total pressure 1.27 atm was collected via water displacement at 29.4 °C, what is the partial pressure of hydrogen gas in the sample? How many grams of aluminum must have reacted to produce this quantity of hydrogen gas? The vapor pressure of water at 29.4 °C is 30.75 torr.
I don't agree with the 22 (22 what?) so I didn't bother with the 54; I assume is wrong also.
To determine the partial pressure of hydrogen gas in the sample, we need to take into account the fact that the collected hydrogen gas is saturated with water vapor.
First, let's convert the given values to SI units:
- Total pressure of the collected gas = 1.27 atm
- Volume of the collected gas = 238.4 mL = 0.2384 L
- Vapor pressure of water at 29.4 °C = 30.75 torr = 0.0405 atm
Next, we can calculate the partial pressure of hydrogen gas by subtracting the vapor pressure of water from the total pressure:
Partial pressure of hydrogen gas = Total pressure - Vapor pressure of water
= 1.27 atm - 0.0405 atm
= 1.23 atm
Therefore, the partial pressure of hydrogen gas in the sample is 1.23 atm.
To determine the number of grams of aluminum that reacted to produce this quantity of hydrogen gas, we can use the ideal gas law. The balanced equation for the reaction between aluminum and hydrogen gas is:
2 Al + 3 H2O -> Al2O3 + 3 H2
The stoichiometry of the reaction tells us that 2 moles of aluminum react to produce 3 moles of hydrogen gas. Thus, we need to find the number of moles of hydrogen gas in the 238.4 mL sample.
We can use the ideal gas law:
PV = nRT
Where:
P = partial pressure of hydrogen gas (in atm)
V = volume of hydrogen gas (in L)
n = number of moles of hydrogen gas
R = ideal gas constant (0.0821 L·atm/K·mol)
T = temperature in Kelvin
Rearranging the equation, we have:
n = PV / RT
Let's calculate the moles of hydrogen gas present in the sample:
n = (1.23 atm) * (0.2384 L) / [(0.0821 L·atm/K·mol) * (29.4 + 273.15) K]
= 0.01324 moles
Since our balanced equation tells us that 2 moles of aluminum react to produce 3 moles of hydrogen gas, we can set up a proportion:
2 moles Al / 3 moles H2 = x moles Al / 0.01324 moles H2
Cross-multiplying and solving for x, we get:
x = (2/3) * 0.01324 moles Al
= 0.00883 moles Al
To calculate the mass of aluminum, we need to know the molar mass of aluminum. It is 26.98 g/mol.
Mass of aluminum = moles of aluminum * molar mass of aluminum
Mass of aluminum = 0.00883 moles * 26.98 g/mol
= 0.238 g
Therefore, approximately 0.238 grams of aluminum must have reacted to produce this quantity of hydrogen gas.
22
54
a.
Ptotal = pH2 + pH2O.
Use PV = nRT and solve for n = number of mols.
2Al + 6H^+ ==> Al^3+ + 3H2
Convert mols H2 to mols Al and to grams.