In the reaction of calcium metal and water, 50.0 mL of hydrogen gas is collected over water at 25.0 degrees C with a total pressure of 752 mmHg. Calcualte the number of moles of hydrogen gas that was produced.
Do we just plug this into n=pv/rt ? or is there like an extra step?
If the total pressure is 752 mm Hg, that is the pressure of the H2 gas + the pressure of the water. So look up the vapor pressure of water at 25 degrees C (at moderate temperatures I remember that the vapor pressure of water is approximately the samae as the T reading in C---but you need to look it up). Subtract. Total P - vapor pressure H2O at 25 C) to give pressure of H2 by itself. Then use PV = nRT to calculate mols H2.
To calculate the number of moles of hydrogen gas produced, you can use the ideal gas law equation, which is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
In this case, you have the pressure (752 mmHg), but it's important to note that the ideal gas law requires the pressure to be in atm units. So, the first step is to convert the pressure to atm. You can do this by dividing it by 760, since 1 atm is equal to 760 mmHg.
752 mmHg / 760 mmHg/atm = 0.9895 atm
Next, you need to convert the volume of the hydrogen gas to liters. Since the gas is collected over water, you need to subtract the partial pressure of water vapor from the total pressure to obtain the pressure of the hydrogen gas alone.
The partial pressure of water vapor at 25.0 degrees C is given as 24.0 mmHg. Subtracting this from the total pressure of 752 mmHg gives:
Pressure of hydrogen gas = Total pressure - Partial pressure of water vapor
= 752 mmHg - 24.0 mmHg
= 728 mmHg
Now, convert the volume of hydrogen gas to liters by dividing the volume of gas collected (50.0 mL) by 1000 to get it in liters:
Volume of hydrogen gas = 50.0 mL / 1000 mL/L
= 0.050 L
Next, make sure to convert the temperature to Kelvin. Add 273.15 to the Celsius temperature to obtain the Kelvin temperature:
Temperature in Kelvin = 25.0 degrees C + 273.15
= 298.15 K
Now you can use the ideal gas law equation to calculate the number of moles of hydrogen gas:
n = PV / RT
n = (0.9895 atm) * (0.050 L) / (0.0821 L·atm/mol·K * 298.15 K)
Simplifying the equation gives:
n ≈ 0.0020 moles
Therefore, approximately 0.0020 moles of hydrogen gas were produced.
To calculate the number of moles of hydrogen gas produced, you can indeed use the ideal gas law equation: n = (P * V) / (R * T), where n represents the number of moles, P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature.
However, in this case, since the measurement of the pressure is given in mmHg and the volume of hydrogen gas is collected over water, you need to apply a correction to the pressure measurement.
When a gas is collected over water, the vapor pressure of water needs to be considered because some of the gas collected will dissolve in the water, affecting the total pressure. The vapor pressure of water at a specific temperature can be obtained from a water vapor pressure table.
In this case, the vapor pressure of water at 25.0 degrees Celsius is approximately 23.8 mmHg. Now you can subtract this vapor pressure of water from the total pressure to get the partial pressure of hydrogen gas.
Partial pressure of hydrogen gas = Total pressure - Vapor pressure of water
Partial pressure of hydrogen gas = 752 mmHg - 23.8 mmHg
Partial pressure of hydrogen gas = 728.2 mmHg
Now you have the corrected pressure value to be used in the ideal gas law equation. Plug in the values into the equation, along with the known values for the ideal gas constant (R = 0.0821 L·atm/(mol·K)) and the volume of the hydrogen gas (50.0 mL = 0.050 L):
n = (P * V) / (R * T)
n = (728.2 mmHg * 0.050 L) / (0.0821 L·atm/(mol·K) * 25.0 + 273.15 K)
After substituting the values and performing the calculation, you will obtain the number of moles of hydrogen gas that was produced in this reaction.