When 25.0 g of Zn reacts, how many L of H2 gas are formed at 25 °C and a pressure of 854 mmHg?
To determine 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 of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, we need to convert the values given to the appropriate units. The pressure is given in mmHg, so we need to convert it to atm. We can use the conversion factor: 1 atm = 760 mmHg.
854 mmHg * (1 atm / 760 mmHg) = 1.1237 atm
Next, we need to convert the temperature from Celsius to Kelvin. We can use the following formula: K = °C + 273.15.
T = 25 °C + 273.15 = 298.15 K
Now, we need to calculate the number of moles of H2 gas produced. We can do this by using the stoichiometry and molar mass of Zn.
1 mole of Zn produces 1 mole of H2 gas according to the balanced chemical equation.
The molar mass of Zn is 65.38 g/mol.
So, for a mass of 25.0 g of Zn, we can calculate the number of moles:
n = (mass / molar mass) = (25.0 g / 65.38 g/mol) = 0.3822 mol
Now, we can substitute the values into the ideal gas law equation:
(1.1237 atm) * V = (0.3822 mol) * (0.0821 L·atm/mol·K) * (298.15 K)
Solving for V:
V = (0.3822 mol) * (0.0821 L·atm/mol·K) * (298.15 K) / (1.1237 atm)
V ≈ 8.437 L
Therefore, when 25.0 g of Zn reacts, approximately 8.437 L of H2 gas are formed at 25 °C and a pressure of 854 mmHg.
To determine the volume of H2 gas formed, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas (854 mmHg)
V = volume of the gas (in L)
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (25°C + 273.15 = 298.15 K)
First, we need to calculate the number of moles of H2 gas formed from the given mass of Zn.
1. Find the molar mass of Zn:
Zn = 65.38 g/mol
2. Calculate the number of moles of Zn:
moles of Zn = mass of Zn / molar mass of Zn
moles of Zn = 25.0 g / 65.38 g/mol
Let's calculate that:
moles of Zn ≈ 0.382 mol
3. Apply stoichiometry to determine the number of moles of H2 produced:
According to the balanced chemical equation for the reaction between Zn and H2:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
The stoichiometric ratio is 1:1 between Zn and H2.
So, the number of moles of H2 produced is also approximately 0.382 mol.
4. Now, we can use the ideal gas law to calculate the volume of H2 gas:
PV = nRT
V = (nRT) / P
Let's plug in the values:
V = (0.382 mol * 0.0821 L·atm/mol·K * 298.15 K) / (854 mmHg)
Now, let's convert mmHg to atm:
1 atm = 760 mmHg
V ≈ (0.382 mol * 0.0821 L·atm/mol·K * 298.15 K) / (854 mmHg * (1 atm / 760 mmHg))
V ≈ 0.043 L
Therefore, approximately 0.043 L of H2 gas is formed.
None if you don't put it with something that will produce H2. I assume it is an acid. 1 mol Zn will produce 1 mole H2.
25.0 g Zn = ?? moles.
moles = grams/molar mass. Then substitute moles Zn you found for n in PV = nRT. Don't forget to use Kelvin for T.