When 25.0g of Zn reacts, how many L of H2 gas are formed at 298K abd a oressure of 854mmHg?
Here is a worked example of a stoichiometry problem. Use that to solve for moles H2, then use PV = nRT t solve for the volume at the conditions listed.
To calculate the number of liters of H2 gas formed, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin
First, we need to calculate the number of moles of Zn. To do that, we divide the given mass of Zn by its molar mass, which is 65.38 g/mol:
Number of moles of Zn = 25.0 g / 65.38 g/mol = 0.382 mol
Since the reaction between Zn and H2 has a 1:1 stoichiometry, the number of moles of H2 gas formed is also 0.382.
Now, we can calculate the volume of H2 gas using the ideal gas law equation:
P = 854 mmHg = 1.124 atm (since 1 atm = 760 mmHg)
T = 298K
V = nRT / P
= (0.382 mol)(0.0821 L.atm/mol.K)(298K) / 1.124 atm
≈ 8.17 L
Therefore, approximately 8.17 liters of H2 gas are formed.
To calculate the volume of H2 gas formed, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L⋅atm/mol⋅K)
T = temperature (in Kelvin)
First, convert the given pressure from mmHg to atm:
1 atm = 760 mmHg
854 mmHg * (1 atm / 760 mmHg) = 1.125 atm
Next, calculate the number of moles of H2 gas formed using the molar mass of Zn and stoichiometry:
1 mole of Zn reacts to produce 1 mole of H2 gas.
Molar mass of Zn = 65.38 g/mol
Number of moles of Zn = mass of Zn / molar mass of Zn
= 25.0 g / 65.38 g/mol
= 0.382 mol
Now, we have the number of moles, pressure, temperature, and the ideal gas constant, so we can substitute these values into the ideal gas law equation to solve for the volume (V):
V = nRT / P
V = (0.382 mol) * (0.0821 L⋅atm/mol⋅K) * (298 K) / (1.125 atm)
V = 8.29 L (rounded to two decimal places)
Therefore, when 25.0 g of Zn reacts at 298 K and a pressure of 854 mmHg, approximately 8.29 L of H2 gas are formed.