I need help calculating the partial pressure of H2 using the following information 0.028 g Mg with excess HCl which produced 31.0 mL of H2 gas. The gas was collected by water displacement in a 22 degrees Celsius water bath. The pressure was 746 mmHg.
To calculate the partial pressure of H2, we need to use the ideal gas law and consider the information provided about the amount of Mg reacted, the volume of H2 gas produced, and the pressure in the water bath.
Here are the steps to calculate the partial pressure of H2 gas:
Step 1: Convert the given mass of Mg to moles.
The molar mass of Mg is 24.31 g/mol. To find the number of moles, we divide the given mass by the molar mass:
0.028 g Mg / 24.31 g/mol Mg = 0.00115 mol Mg
Step 2: Use the balanced chemical equation to find the moles of H2 produced.
The balanced equation for the reaction between Mg and HCl is:
Mg + 2HCl -> MgCl2 + H2
From the equation, we can see that 1 mole of Mg reacts to produce 1 mole of H2. Therefore, the number of moles of H2 produced is also 0.00115 mol.
Step 3: Use the ideal gas law equation to find the partial pressure of H2.
The ideal gas law equation is:
PV = nRT
Where:
P is the pressure (in atmospheres)
V is the volume (in liters)
n is the number of moles
R is the ideal gas constant (0.0821 L.atm/mol.K)
T is the temperature (in Kelvin)
First, we need to convert the volume of H2 gas collected to liters. Since the volume of H2 gas is given in milliliters, we divide it by 1000:
31.0 mL / 1000 mL/L = 0.0310 L
Next, we need to convert the temperature to Kelvin. The given temperature is 22 degrees Celsius, so we add 273.15 to it:
22 °C + 273.15 = 295.15 K
Now, we can plug the values into the ideal gas law equation:
P(H2) * V(H2) = n(H2) * R * T
Let's rearrange the equation to solve for the partial pressure of H2 (P(H2)):
P(H2) = (n(H2) * R * T) / V(H2)
P(H2) = (0.00115 mol * 0.0821 L.atm/mol.K * 295.15 K) / 0.0310 L
P(H2) ≈ 6.77 atm
Step 4: Convert the pressure from atm to mmHg.
To convert the pressure from atm to mmHg, we multiply it by 760, since 1 atm is equivalent to 760 mmHg:
P(H2) ≈ 6.77 atm * 760 mmHg/atm
P(H2) ≈ 5145.2 mmHg
Therefore, the partial pressure of H2 gas is approximately 5145.2 mmHg.