zinc is mixed with Hydrochloric acid to form zinc chloride and hydrogen gas. if 7.3 grams of zinc completely reacts how much hydrogen gas will be created? if you collect the hydrogen over water at 23 degrees celcius and a barometric pressure of 770 mmHg and you recover 2.30 liters of H2 what is the percent yield? how much H2 did you recover adjusted to stp ?
To find out how much hydrogen gas will be created when 7.3 grams of zinc completely reacts with hydrochloric acid, we need to refer to the balanced chemical equation for the reaction:
Zn + 2HCl -> ZnCl2 + H2
According to the equation, 1 mole of zinc reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas. Therefore, we need to convert the mass of zinc to moles to determine the amount of hydrogen gas produced.
1. Calculate the molar mass of zinc (Zn):
Molar mass of Zn = 65.38 g/mol
2. Convert the mass of zinc to moles:
Moles of Zn = Mass of Zn / Molar mass of Zn
Moles of Zn = 7.3 g / 65.38 g/mol
Moles of Zn ≈ 0.1113 moles of Zn
Since the stoichiometric ratio between Zn and H2 is 1:1, the moles of hydrogen gas produced will be the same as the moles of zinc.
Moles of H2 = Moles of Zn ≈ 0.1113 moles of H2
To calculate the amount of hydrogen gas collected over water at 23 degrees Celsius and 770 mmHg, we need to use the ideal gas law equation:
PV = nRT
where:
P = pressure in atm
V = volume in liters
n = moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
3. Convert the barometric pressure to atm:
Barometric pressure = 770 mmHg
Barometric pressure in atm = 770 mmHg / 760 mmHg/atm
Barometric pressure in atm ≈ 1.0132 atm
4. Convert the temperature from Celsius to Kelvin:
Temperature in Celsius = 23 degrees Celsius
Temperature in Kelvin = 23 + 273.15
Temperature in Kelvin ≈ 296.15 K
5. Use the ideal gas law equation to calculate the moles of H2 at the given conditions:
n = PV / RT
Moles of H2 = (1.0132 atm * 2.30 L) / (0.0821 L·atm/mol·K * 296.15 K)
Moles of H2 ≈ 0.0932 moles of H2
The moles of hydrogen gas collected can be considered as the theoretical yield.
To calculate the percent yield, we use the equation:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
6. Calculate the percent yield:
Percent Yield = (0.0932 moles / 0.1113 moles) * 100
Percent Yield ≈ 83.7%
Lastly, to calculate the amount of hydrogen gas recovered at STP (standard temperature and pressure: 0 degrees Celsius and 1 atm), we can use the ideal gas law:
PV = nRT
where:
P = 1 atm
V = volume in liters
n = moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
7. Convert the temperature from Celsius to Kelvin:
Temperature in Celsius = 0 degrees Celsius
Temperature in Kelvin = 0 + 273.15
Temperature in Kelvin ≈ 273.15 K
8. Use the ideal gas law equation to calculate the volume of H2 at STP:
V = nRT / P
Volume of H2 at STP = (0.0932 moles * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm
Volume of H2 at STP ≈ 2.12 liters
Therefore, the percent yield is approximately 83.7%, and the amount of hydrogen gas recovered at STP is approximately 2.12 liters.
To determine the amount of hydrogen gas produced when 7.3 grams of zinc completely reacts, you need to use stoichiometry.
First, write and balance the chemical equation for the reaction:
Zn + 2HCl -> ZnCl2 + H2
From the balanced equation, you can see that the ratio between zinc and hydrogen is 1:1. This means that for every 1 mole of zinc reacted, 1 mole of hydrogen gas is produced.
1 mole of zinc (Zn) is equal to its molar mass (65.38 g) and 1 mole of hydrogen gas (H2) is equal to its molar mass (2.02 g).
Now, calculate the number of moles of zinc:
moles of Zn = mass of Zn / molar mass of Zn
moles of Zn = 7.3 g / 65.38 g/mol ≈ 0.1114 mol
Since the reaction is 1:1, the number of moles of hydrogen gas produced is also 0.1114 mol.
To determine the volume of hydrogen gas collected over water at the given conditions, we need to apply the ideal gas law:
PV = nRT
Where:
P = pressure (converted to atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature (converted to Kelvin)
First, convert the pressure from mmHg to atm:
770 mmHg / 760 mmHg/atm ≈ 1.0132 atm
Next, convert the temperature from Celsius to Kelvin:
23 degrees Celsius + 273.15 ≈ 296.15 K
Now, rearrange the ideal gas law equation to solve for volume (V):
V = (nRT) / P
V = (0.1114 mol * 0.0821 L·atm/(mol·K) * 296.15 K) / 1.0132 atm ≈ 2.03 L
Therefore, the volume of hydrogen gas collected over water is approximately 2.03 liters.
To determine the percentage yield, we need to compare the actual yield (amount of H2 collected) to the theoretical yield (amount of H2 calculated from stoichiometry).
To find the theoretical yield of hydrogen:
Theoretical yield (in grams) = moles of H2 * molar mass of H2
Theoretical yield = 0.1114 mol * 2.02 g/mol ≈ 0.225 g
To calculate the percentage yield:
Percentage yield = (actual yield / theoretical yield) * 100%
Percentage yield = (2.30 L / 0.225 g) * 100% ≈ 1022.22%
The percent yield is approximately 1022.22%.
To determine the volume of hydrogen gas collected at STP (Standard Temperature and Pressure), you need to use the ideal gas law.
The temperature at STP is 273.15 K, and the pressure is 1 atm.
V = nRT / P
V = (0.1114 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
V ≈ 2.31 L
Therefore, the volume of hydrogen gas collected adjusted to STP is approximately 2.31 liters.