0.75 L of hydrogen gas is collected over water at 25 oC and 101.6. What volumme will e dry hydrogen occupy 103.3 kPa and 25 oC?
What is the balanced equation and what of numbers do i use ?
To determine the volume of dry hydrogen gas at a different temperature and pressure, you can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin (K)
First, let's convert the given values to the appropriate units:
P1 = 101.6 kPa (this is the pressure at 25 °C)
T1 = 25 °C = 298 K
Now, we need to find the number of moles of hydrogen gas (n1) using the ideal gas law equation. Rearranging the equation, we have:
n1 = (P1 * V1) / (R * T1)
Here, V1 is the volume of hydrogen gas at the first set of conditions, which is 0.75 L. Plugging in the values:
n1 = (101.6 kPa * 0.75 L) / (8.314 J/(mol·K) * 298 K)
Note: Make sure to use the appropriate gas constant units to match the units in the ideal gas law equation.
Now, we can calculate the number of moles of hydrogen gas.
Next, we will use the molar ratio of the balanced chemical equation to determine the stoichiometry of the reaction. Unfortunately, you have not provided the reaction equation, so I cannot provide the specific molar ratio or balanced equation at the moment.
Once you have the balanced equation, identify the stoichiometric relationship between the reactant and product. This will allow you to determine the number of moles of hydrogen gas (n2) at the new set of conditions.
Finally, once you have n2, rearrange the ideal gas law equation and solve for V2:
V2 = (n2 * R * T2) / P2
Here, T2 is the new temperature in Kelvin, which is also 25 °C or 298 K. P2 is the new pressure, which is 103.3 kPa. Plugging in the values, you can calculate the volume of dry hydrogen gas.
101.6 WHAT? kPa?
Use (P1V1) = (P2V2)
P1 = 101.6 kPa - vapor pressure H2O (in kPa).
V1 = 0.75
P2 = 101.3 kPa
V2 = solve for this