Zn + 2 HCl = H2 + ZnCl2 / When 25.0 g of Zn reacts, how many L of H2 gas are formed at 25 degrees Celsius and a pressure of 854 mmHg?
mols Zn = grams/atomic mass Zn = ?
mols H2 formed = same as mols Zn (look at the 1 mol Zn forming 1 mol H2)
Substitute mols H2 into PV = nRT at the conditions listed and solve for volume in L. Remember T must be in kelvin and P in atmospheres for V to come out in L. Use 0.08206 L*atm/mol*K for R
To determine the volume of H2 gas produced, we need to use the Ideal Gas Law equation: PV = nRT.
Given:
Mass of Zn = 25.0 g
Temperature (T) = 25 degrees Celsius = 25 + 273 = 298 K
Pressure (P) = 854 mmHg
First, we need to convert the mass of Zn to moles using its molar mass.
The molar mass of Zn = 65.38 g/mol
Moles of Zn = Mass of Zn / Molar mass of Zn
Moles of Zn = 25.0 g / 65.38 g/mol
Moles of Zn = 0.382 mol
From the balanced chemical equation, we can see that the stoichiometric ratio between moles of Zn and moles of H2 is 1:1. Therefore, the moles of H2 gas produced will also be 0.382 mol.
Now, we can use the ideal gas law to find the volume (V) of H2 gas.
PV = nRT
R is the ideal gas constant, which is 0.0821 L·atm/(mol·K)
V = (nRT) / P
V = (0.382 mol * 0.0821 L·atm/(mol·K) * 298 K) / 854 mmHg
Note: For calculations, it is important to use consistent units. Since the pressure is given in mmHg, we should convert it to atm.
Now, we can convert mmHg to atm:
1 atm = 760 mmHg
Pressure (P) = 854 mmHg / 760 mmHg/atm = 1.1237 atm
V = (0.382 mol * 0.0821 L·atm/(mol·K) * 298 K) / 1.1237 atm
V ≈ 8.24 L
Therefore, approximately 8.24 liters of H2 gas are formed at 25 degrees Celsius and a pressure of 854 mmHg.
To solve this problem, you need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's calculate the number of moles of Zn. To do that, you need to know the molar mass of Zn, which is 65.38 g/mol. To find the number of moles, divide the given mass by the molar mass:
Number of moles = Mass / Molar mass
Number of moles = 25.0 g / 65.38 g/mol ≈ 0.382 moles
Now, let's use the balanced chemical equation to determine the stoichiometry between Zn and H2. From the reaction, we can see that for each mole of Zn, one mole of H2 is produced.
So, the number of moles of H2 gas formed is also 0.382 moles.
Next, convert the given temperature from Celsius to Kelvin by adding 273.15:
T = 25 degrees Celsius + 273.15 = 298.15 K
Now, let's plug the values into the ideal gas law equation:
PV = nRT
We need to find the volume of H2 gas (V), so rearrange the equation:
V = (nRT) / P
Plugging in the values:
V = (0.382 moles × 0.0821 L·atm/mol·K × 298.15 K) / 854 mmHg
Now, we need to convert mmHg to atm because the ideal gas constant is in atm units.
1 atm = 760 mmHg
V = (0.382 moles × 0.0821 L·atm/mol·K × 298.15 K) / (854 mmHg / 760 mmHg)
V = (0.382 moles × 0.0821 L·atm/mol·K × 298.15 K) / 1.121
V ≈ 8.085 L
Therefore, approximately 8.085 L of H2 gas will be formed when 25.0 g of Zn reacts at 25 degrees Celsius and a pressure of 854 mmHg.