How many grams of hydrogen are collected in a reaction where 1.80L of hydrogen gas is collected over water at a temperature of 40degrees C and a total pressure of 738torr?
Use PV = nRT
Remember T must be in kelvin.
Look up vapor pressure H2O at 40 C.
Substitute [(738-vpH2O)/760] for P and solve for n = number of mols. Then n = grams/molar mass. You know n and molar mass, solve for grams.
To find the number of grams of hydrogen collected in the given reaction, we start by using the ideal gas law equation: PV = nRT.
First, we need to calculate the pressure of the hydrogen gas alone. Since the hydrogen gas is collected over water, the partial pressure of the hydrogen gas is equal to the total pressure of the collected gas minus the vapor pressure of water at the given temperature.
Next, we convert the given volume of hydrogen gas to liters, as required by the ideal gas law equation.
Finally, we rearrange the ideal gas law equation to solve for the number of moles of hydrogen gas, and then use the molar mass of hydrogen to convert moles to grams.
Let's break down the steps:
Step 1: Calculate the partial pressure of hydrogen gas.
The vapor pressure of water at 40 degrees Celsius is 55.3 torr, so:
Partial pressure of hydrogen gas = Total pressure of the collected gas - Vapor pressure of water
Partial pressure of hydrogen gas = 738 torr - 55.3 torr
Partial pressure of hydrogen gas = 682.7 torr
Step 2: Convert the given volume of hydrogen gas to liters.
1.80 L (given volume of hydrogen gas)
Step 3: Rearrange the ideal gas law equation to solve for the number of moles of hydrogen gas.
PV = nRT
n (moles of gas) = PV / RT
R is the ideal gas constant, which is 0.0821 L·atm/mol·K.
T is the temperature in Kelvin, so we need to convert 40 degrees Celsius to Kelvin:
T (Kelvin) = 40 + 273.15 = 313.15 K
n (moles of hydrogen gas) = (Partial pressure of hydrogen gas in atm) * (Volume of hydrogen gas in liters) / (R constant * Temperature in Kelvin)
Step 4: Convert moles of hydrogen gas to grams.
We know that 1 mole of hydrogen gas has a molar mass of 2 grams.
Grams of hydrogen gas = (moles of hydrogen gas) * (molar mass of hydrogen)
Now, let's calculate:
Partial pressure of hydrogen gas (in atm) = 682.7 torr / 760 torr/atm = 0.897 atm
n (moles of hydrogen gas) = (0.897 atm) * (1.80 L) / (0.0821 L·atm/mol·K * 313.15 K)
n (moles of hydrogen gas) ≈ 0.0818 mol
Grams of hydrogen gas = 0.0818 mol * 2 g/mol
Grams of hydrogen gas ≈ 0.164 g
Therefore, approximately 0.164 grams of hydrogen gas are collected in the given reaction.