Ca(s) + 2H+(aq) → Ca2+(aq) + H2(g)
An experiment is performed to determine the volume of hydrogen gas produced when calcium reacts with excess HCl(aq) as represented by the net ionic equation above. A 0.0720 g sample of pure calcium and a solution of 1.0 M HCl(aq) were used in the experiment.
(a) Calculate the number of moles of calcium in the 0.0720 g sample.
My answer: 0.001796 moles Ca
(b) Calculate the number of moles of HCl(aq) needed to react completely with the sample of calcium.
My answer: 0.003592 moles HCl
As the calcium reacts, the hydrogen gas produced is collected by water displacement at 25°C. The pressure of the gas in the collection tube is measured to be 752 torr.
(c) Given that the equilibrium vapor pressure of water is 24 torr at 25.0°C, calculate the pressure that the H2(g)
produced in the reaction would have if it were dry.
(d) Calculate the dry volume, in liters measured at the conditions in the laboratory, that the H2(g) would have
produced in the reaction.
(e) The laboratory procedure specified 1.0 M HCl solution be, but only 6.0 M HCl solution was available. Describe
the steps for safely preparing 50.0 mL of 1.0 M HCl(aq) using 6.0 M HCl solution and materials selected from
the list below. Show any necessary calculation(s).
10.0 mL graduated cylinder
Distilled water
250 mL beakers
Balance
50.00 mL volumetric flask
Dropper
For parts (c) and (d), when it says "dry," does that mean ignore the presence of the water vapor? So for (c), calculate the pressure of the gas without the equilibrium vapor pressure of water? Would you just subtract 24 torr from 752 torr? For (d), how does you find the "dry volume"?
For part (e), how much water would you add to distill the 6.0M HCl to 1.0M HCl?
c you have right but for the wrong reason. You don't IGNORE the water vapor, you count it.
Ptotal = pH2 + pH2O
pH2 = Ptotal - pH2O
For part e, you don't distill anything. You dilute 6.0M HCl so its final concn is 1.0M.
mL1 x M1 = mL2 x M2
50 mL x 1.0M = mL2 x 6.0
mL2 = 50 x 1/6 = about 8.33
So you use the graduated cylinder to measure out 8.3 mL of the 6.0M HCl, pour that into the 50 mL volumetric flask, add distilled H2O to the mark in the flask, and mix thoroughly.
I don't know if this is accurate enough for your prof; I don't know how accurate you are to be. If you need more accuracy than you can get as described above, you can calibrate the dropper and add the appropriate number of drops to obtain 8.33 mL (or at least as close as that will permit).
For part (c), "dry" means that you should consider the pressure of the hydrogen gas without taking into account the presence of water vapor. To calculate the pressure of the dry hydrogen gas, you would subtract the equilibrium vapor pressure of water (24 torr) from the measured pressure of the gas in the collection tube (752 torr).
Pressure of H2(g) (dry) = Pressure of gas in collection tube - Equilibrium vapor pressure of water
= 752 torr - 24 torr
= 728 torr
For part (d), the "dry volume" refers to the volume of the hydrogen gas that would have been produced in the reaction if it were completely dry, without considering the presence of water vapor. To calculate the dry volume, you need to apply the Ideal Gas Law equation: PV = nRT.
First, we need to convert the pressure from torr to atm:
Pressure of H2(g) (dry) = 728 torr / 760 torr/atm
≈ 0.958 atm
Next, we need to calculate the number of moles of hydrogen gas produced in the reaction. From the balanced equation, we know that the ratio of moles of calcium to moles of hydrogen gas is 1:1. Therefore, the number of moles of hydrogen gas is equal to the number of moles of calcium (0.001796 moles).
Now we can calculate the dry volume using the ideal gas law equation:
PV = nRT
where:
P = Pressure in atm (0.958 atm)
V = Volume in liters (unknown)
n = number of moles (0.001796 moles)
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature in Kelvin (25°C = 298.15 K)
Rearranging the equation:
V = nRT / P
Substituting the values:
V = (0.001796 mol)(0.0821 L·atm/mol·K)(298.15 K) / (0.958 atm)
≈ 0.0466 L or 46.6 mL
So, the dry volume of the hydrogen gas produced in the reaction would be approximately 46.6 mL.
For part (e), to prepare 50.0 mL of 1.0 M HCl(aq) using a 6.0 M HCl solution, you can use the following steps:
1. Determine the volume of the 6.0 M HCl solution needed to obtain a 1.0 M HCl solution volume:
(1.0 M)(50.0 mL) = (6.0 M)(V)
V ≈ 8.33 mL
2. Use the graduated cylinder to measure 8.33 mL of the 6.0 M HCl solution.
3. Pour the measured 6.0 M HCl solution into a 50.0 mL volumetric flask.
4. Add distilled water to the volumetric flask up to the 50.0 mL mark.
5. Mix the solution thoroughly by swirling or inverting the flask, ensuring complete mixing.
By following these steps, you will be able to prepare 50.0 mL of 1.0 M HCl(aq) using the 6.0 M HCl solution and the available materials.
For part (c), when it says "dry," it means to calculate the pressure of the hydrogen gas without considering the presence of the water vapor. To find the pressure of the dry hydrogen gas, you subtract the equilibrium vapor pressure of water from the measured pressure of the gas. So in this case, you would subtract 24 torr from the measured pressure of 752 torr to get the pressure of the dry hydrogen gas.
For part (d), to find the dry volume of the hydrogen gas, you need to use the ideal gas law equation: PV = nRT. Since you already have the number of moles of the hydrogen gas (which is equal to the number of moles of calcium reacted in this case), you can rearrange the equation to solve for volume:
V = nRT/P
Where:
V = volume of gas
n = number of moles of gas
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature in Kelvin
P = pressure of the gas
In this case, since the temperature is not given, assume it is 25°C which is equivalent to 298 K. You would plug in the values for n (0.001796 moles), R, T, and the pressure of the dry hydrogen gas (from part (c)) into the equation to calculate the dry volume of the hydrogen gas.
For part (e), to prepare 50.0 mL of 1.0 M HCl(aq) using 6.0 M HCl solution, you need to dilute the 6.0 M HCl solution with distilled water.
The formula to calculate the dilution is:
C1V1 = C2V2
Where:
C1 = initial concentration of the HCl solution (6.0 M)
V1 = initial volume of the HCl solution (unknown)
C2 = final concentration of the HCl solution (1.0 M)
V2 = final volume of the HCl solution (50.0 mL)
To solve for V1, rearrange the equation:
V1 = (C2V2) / C1
Plug in the values C1 (6.0 M), C2 (1.0 M), and V2 (50.0 mL) into the equation to calculate the volume of the 6.0 M HCl solution you need to mix with distilled water. Use a 10.0 mL graduated cylinder to measure out 10.0 mL of the 6.0 M HCl solution and transfer it to a 250 mL beaker. Then use a dropper to slowly add distilled water to the beaker while stirring until you reach the desired final volume of 50.0 mL in a 50.00 mL volumetric flask.