Suppose we collect hydrogen gas, H2(g), over water at 21◦C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express
your answer in liters.
2Al + 6HCl => 2AlCl3 + 3H2
(1.35/27)mole Al = 0.05molAl => 3/2(.05)mol H2 = 0.075mol H2
Vol H2 = nRT/P = [(0.075)(0.08206)(294)(760)]/(743) = 1.85L H2
1.9L
To find the volume of hydrogen gas collected, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
First, we need to calculate the number of moles of hydrogen gas produced from the reaction between aluminum and hydrochloric acid.
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
From the balanced equation, we can see that 1 mole of aluminum (Al) reacts to produce 3 moles of hydrogen gas (H2).
Using the molar mass of aluminum, we can calculate the number of moles of aluminum:
Molar mass of Al = 26.98 g/mol
Mass of Al = 1.35 g
Number of moles of Al = Mass of Al / Molar mass of Al
= 1.35 g / 26.98 g/mol
Next, we can use the mole ratio between aluminum and hydrogen gas to find the number of moles of hydrogen gas produced:
Moles of H2 = Moles of Al * (3 moles H2 / 2 moles Al)
Now, let's convert the temperature from Celsius to Kelvin:
Temperature in Kelvin = 21 + 273 = 294 K
Now we can substitute the values into the ideal gas law equation to find the volume of hydrogen gas:
PV = nRT
V = (nRT) / P
Plug in the values:
V = (moles of H2 * 0.0821 L·atm/mol·K * 294 K) / 743 Torr
Note: We need to convert Torr to atm by dividing by 760 Torr/atm.
V = (moles of H2 * 0.0821 L·atm/mol·K * 294 K) / (743 Torr / 760 Torr/atm)
Now, we can calculate the volume:
V = (moles of H2 * 0.0821 L·atm/mol·K * 294 K * 760 Torr / 743 Torr)
Finally, substitute the calculated value of moles of H2 into the equation to find the volume of hydrogen gas collected in liters.
To find the volume of hydrogen gas collected, we need to use the ideal gas law equation:
PV = nRT
where:
P is the pressure of the gas,
V is the volume of the gas,
n is the number of moles of gas,
R is the ideal gas constant (0.0821 L·atm/(mol·K)), and
T is the temperature of the gas in Kelvin.
First, we need to calculate the number of moles of hydrogen gas produced, given the mass of aluminum reacted.
1. Calculate the moles of aluminum (Al) using its molar mass:
Molar mass of Al = 26.98 g/mol
Moles of Al = mass of Al / molar mass of Al
Moles of Al = 1.35 g / 26.98 g/mol
2. Determine the ratio of moles of hydrogen (H2) to moles of aluminum (Al) from the balanced equation:
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
From the balanced equation, 2 moles of Al react to produce 3 moles of H2.
3. Calculate the moles of hydrogen gas produced:
Moles of H2 = (Moles of Al) × (3 moles of H2 / 2 moles of Al)
Now, we have the number of moles of hydrogen gas produced.
Next, we can use the ideal gas law equation to find the volume of the gas.
4. Convert the temperature from Celsius to Kelvin:
Temperature in Kelvin = 21°C + 273.15
5. Rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
Plug in the values and solve for the volume (V) in liters.
Now, let's perform the calculations step by step:
Step 1: Calculate moles of aluminum (Al)
Molar mass of Al = 26.98 g/mol
Moles of Al = 1.35 g / 26.98 g/mol = 0.05002 mol
Step 2: Determine the ratio of moles of hydrogen (H2) to moles of aluminum (Al)
From the balanced equation: 2 moles of Al react to produce 3 moles of H2.
Step 3: Calculate moles of hydrogen gas produced
Moles of H2 = (0.05002 mol) × (3 moles of H2 / 2 moles of Al) = 0.07503 mol
Step 4: Convert temperature from Celsius to Kelvin
Temperature in Kelvin = 21°C + 273.15 = 294.15 K
Step 5: Use the ideal gas law equation to find the volume
V = (nRT) / P
V = (0.07503 mol) × (0.0821 L·atm/(mol·K)) × (294.15 K) / 743 Torr
743 Torr is equal to 743/760 atm (since 1 atm = 760 Torr)
V = (0.07503 mol) × (0.0821 L·atm/(mol·K)) × (294.15 K) / (743/760 atm)
V ≈ 0.0585 L
So, the volume of hydrogen gas collected will be approximately 0.0585 liters.