472 mL of H2 was collected over water when 1.256g Zn reacted with excess HCl. The atmospheric pressure during the experiment was 754 mmHg and the temp was 26C. What is the pressure in atm of the dry hydrogen gas in the mixture.
I don't know how to answer this.
The way the problem is stated makes it appear there is a lot of information that isn't used. Most of these problems want to know the pressure of the dry H2 gas at STP; however, this one just asks for pressure of dry H2 gas so I assume that is pressure at 25 C but dry gas instead of wet. That simplifies the problem quite a bit.
Ptotal = pH2 dry + pH2O
754 mm Hg = pH2 + pH2O.
Look up pH2O in your text/notes. It is a function of T and it will be called "vapor pressure" tables or vapor pressure H2O. I don't have these tables memorized but at 26 C the vapor pressure of H2O will be approx 25 mm.
Substitute that into the formula above and solve for pH2. That will be the pressure of dry H2 at 26 C. The problem asks for P in atm so convert. P in atm = P in mm/760 = ?
To answer this question, you need to apply Dalton's law of partial pressures and take into account the vapor pressure of water at the given temperature. Here's how you can calculate the pressure of the dry hydrogen gas:
1. Convert the volume of hydrogen collected over water to its standard volume at STP (standard temperature and pressure). Since the question provides the temperature and the atmospheric pressure during the experiment, you can use the ideal gas law equation: PV = nRT.
- Convert the given temperature of 26°C to Kelvin by adding 273: 26°C + 273 = 299 K.
- Convert the given atmospheric pressure of 754 mmHg to atm by dividing it by 760 (1 atm = 760 mmHg): 754 mmHg ÷ 760 = 0.992 atm.
Now you have the temperature (299 K) and the pressure (0.992 atm). Plug these values along with the volume (472 mL) into the ideal gas law equation as follows:
(0.992 atm) * (V mL) = (n moles) * (0.0821 L·atm·K⁻¹·mol⁻¹) * (299 K)
2. Convert the volume of hydrogen gas to liters:
472 mL ÷ 1000 = 0.472 L
3. Convert the mass of zinc (1.256 g) to moles. Calculate the molar mass of zinc (Zn) from the periodic table (65.38 g/mol):
1.256 g ÷ (65.38 g/mol) = 0.0192 mol Zn
4. According to the balanced chemical equation, the reaction between zinc (Zn) and hydrochloric acid (HCl) produces hydrogen gas (H2) in a 1:1 ratio. Therefore, the number of moles of hydrogen gas collected is also 0.0192 mol.
Now, rearrange the ideal gas law equation to solve for volume:
(0.992 atm) * (0.472 L) = (0.0192 mol) * (0.0821 L·atm·K⁻¹·mol⁻¹) * (299 K)
5. Solve for the volume of the dry hydrogen gas (V mL):
V mL = (0.0192 mol * 0.0821 L·atm·K⁻¹·mol⁻¹ * 299 K) / (0.992 atm)
Plug in the given values and calculate V mL.
6. The calculated volume (V mL) represents the volume of the wet hydrogen gas. To determine the volume of the dry hydrogen gas, you need to subtract the amount of water vapor present.
To calculate the vapor pressure of water at 26°C, consult a water vapor pressure table or use the Antoine equation. For simplicity, let's assume the vapor pressure of water at 26°C is 25.2 mmHg.
Convert the vapor pressure to atm: 25.2 mmHg ÷ 760 = 0.033 atm.
Subtract the vapor pressure of water from the atmospheric pressure to get the pressure of the dry hydrogen gas:
Pressure of dry H2 gas = Atmospheric pressure - Vapor pressure of water
Pressure of dry H2 gas = 0.992 atm - 0.033 atm
7. Calculate and report the pressure of the dry hydrogen gas.
So, the pressure of the dry hydrogen gas in the mixture is 0.959 atm.