What volume of H2 gas (in liters), measured at 23 °C and 678 torr, can be obtained by reacting 7.59 g of zinc metal with 124 mL of 0.302 M HCl?
Zn + 2HCl --> ZnCl2 + H2
This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.
mols Zn = grams/atomic mass = ?
mols HCl = M x L = ?
Using the coefficients in the balanced equation, convert mols Zn from above into mols H2.
Do the same for mols HCl to mols H2.
It is likely that these two values will not be the which means one is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Using the smaller value for mols, use PV = nRT to solve for V in liters.
To solve this problem, we need to use the ideal gas law equation, which is given as follows:
PV = nRT
Where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles of the gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
To find the volume of H2 gas, we need to determine the number of moles of H2 gas produced.
Step 1: Convert the temperature from Celsius to Kelvin.
Given: T = 23 °C
To convert to Kelvin, we use the formula:
T(K) = T(°C) + 273.15
T(K) = 23 + 273.15
T(K) = 296.15K
Step 2: Convert the pressure from torr to atm.
Given: P = 678 torr
To convert to atm, we use the conversion factor:
1 atm = 760 torr
P(atm) = P(torr) / 760
P(atm) = 678 / 760
P(atm) = 0.8921 atm
Step 3: Calculate the number of moles of H2 gas produced.
Since the stoichiometry of the balanced equation tells us that 1 mole of zinc metal produces 1 mole of H2 gas, we need to calculate the number of moles of zinc metal first.
Given: Mass of zinc (Zn) = 7.59 g
Atomic mass of zinc (Zn) = 65.38 g/mol (from the periodic table)
n(Zn) = Mass(Zn) / Molar mass(Zn)
n(Zn) = 7.59 g / 65.38 g/mol
n(Zn) ≈ 0.116 mol
Since 1 mole of zinc (Zn) produces 1 mole of H2 gas, the number of moles of H2 gas produced is also 0.116 mol.
Step 4: Substitute the given values into the ideal gas law equation to find the volume (V) of H2 gas.
PV = nRT
V = (nRT) / P
Substituting the known values:
V = (0.116 mol * 0.0821 L.atm/mol.K * 296.15K) / 0.8921 atm
V ≈ 3.85 L
Therefore, the volume of H2 gas obtained is approximately 3.85 liters.