A piece of solid magnesium is reacted with hydrochloric acid to form hydrogen gas:
Mg (s) + 2HCl (aq) ---> MgCl2 (aq) + H2 (g)
What volume of hydrogen gas (in mL) is collected over water at 25 degrees Celsius by reaction of 0.450 g of Mg (AW= 24.3 g/mol) with 50.0 mL of 1.0 M HCl? The barometer records an atmospheric pressure of 758 torr and the vapor pressure of water at this temperature is 23.35 torr.
Mg + 2HCl ==> MgCl2 + H2
mols Mg = 1.00 mol
mol HCl = 0.5M
mols H2 produced from 1. mol Mg + 1 mol
mols H2 produced from 0.5 mol HCl = 1 mol
So this isn't a LR problem after all but the reactants just match. So 1 mol H2 is produced and at STP 1 mol occupies 22.4 L
To find the volume of hydrogen gas collected over water, we need to consider the pressure of the hydrogen gas. The total pressure of the system is equal to the sum of the atmospheric pressure and the vapor pressure of water.
Step 1: Calculate the number of moles of Mg used.
Given:
Mass of Mg = 0.450 g
Atomic weight of Mg (AW) = 24.3 g/mol
Number of moles of Mg = mass of Mg / AW of Mg
Number of moles of Mg = 0.450 g / 24.3 g/mol
Step 2: Calculate the total pressure.
Total Pressure = atmospheric pressure + vapor pressure of water
Total Pressure = 758 torr + 23.35 torr
Step 3: Calculate the moles of H2 gas produced.
From the balanced equation, we know that 1 mole of Mg reacts to produce 1 mole of H2 gas. So the moles of H2 gas produced will be equal to the moles of Mg used.
Moles of H2 gas = Moles of Mg
Step 4: Calculate the volume of H2 gas collected.
Using the ideal gas law equation: PV = nRT
V = (nRT) / P
Where:
V = volume of H2 gas
n = moles of H2 gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
P = total pressure
Conversion:
Temperature in Celsius to Kelvin: K = °C + 273.15
Step 5: Convert the volume from liters to milliliters.
1 liter (L) = 1000 milliliters (mL)
Let's plug in the values and calculate the volume of hydrogen gas collected over water.
Moles of Mg = 0.450 g / 24.3 g/mol
Total Pressure = 758 torr + 23.35 torr
T = 25°C + 273.15
R = 0.0821 L·atm/mol·K
V = [(0.450 g / 24.3 g/mol) * (0.0821 L·atm/mol·K) * (25°C + 273.15 K)] / (758 torr + 23.35 torr)
V = [(0.450 / 24.3) * (0.0821) * (25 + 273.15)] / (758 + 23.35)
V = 0.0126 L
Now, let's convert the volume from liters to milliliters:
V (mL) = 0.0126 L * 1000 mL/L
V (mL) = 12.6 mL
Therefore, the volume of hydrogen gas collected over water is 12.6 mL.