4. An experiment similar to yours was done by Julio. He used 0.0024 mole of lead instead of magnesium. He collected 45 ml of wet hydrogen gas at 25 degrees Celsius and 0.98 atm. How many ml of H2 gas should Julio have collected? What was Julio’s percent error?
To determine the number of milliliters of H2 gas Julio should have collected, we need to use the ideal gas equation, PV = nRT, where:
P = pressure in atm
V = volume in liters
n = moles of gas
R = ideal gas constant (0.0821 atm·L/mol·K)
T = temperature in Kelvin
First, let's convert the given conditions:
Pressure, P = 0.98 atm
Volume, V = 45 mL = 0.045 L (converted from mL to L)
Temperature, T = 25 degrees Celsius = 25 + 273 = 298 K (converted from Celsius to Kelvin)
Now, let's calculate the moles of hydrogen gas based on the moles of lead used, assuming the reaction is balanced:
1 Pb + 2 HCl -> 1 PbCl2 + 1 H2
From the balanced equation, we see that 1 mole of lead produces 1 mole of H2. Therefore, if Julio used 0.0024 moles of lead, he should have produced 0.0024 moles of H2 gas.
Now, we can plug these values into the ideal gas equation:
PV = nRT
(0.98 atm) * (V) = (0.0024 mol) * (0.0821 atm·L/mol·K) * (298 K)
Solve for V:
V = (0.0024 mol * 0.0821 atm·L/mol·K * 298 K) / (0.98 atm)
V ≈ 0.059 L
Finally, let's convert the volume back to milliliters:
0.059 L * 1000 mL/L = 59 mL
Therefore, Julio should have collected 59 mL of H2 gas.
To calculate Julio's percent error, we need to compare the actual value (59 mL) to the given value (45 mL):
Percent error = (|Experimental Value - Accepted Value| / Accepted Value) * 100
Percent error = (|59 mL - 45 mL| / 45 mL) * 100
Percent error = (14 mL / 45 mL) * 100 ≈ 31.1%
Therefore, Julio's percent error is approximately 31.1%.