Acetylene(C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reacts with water: CaC2(s) + 2H20(l) → C2H2(g) + Ca(OH)2(aq) For a sample of acetylene collected over water, total gas pressure (adjusted to barometric pressure) is 738 torr and the volume is 523 mL. At the temperature of the gas(23∘C), the vapor pressure of water is 2 1 torr. How many grams of acetylene are collected?
NOTE: I don't believe the vapor pressure of H2O at 23 C is 2.1 torr. It's more like 21 torr. I think you made a typo or it's wrong in the problem.
Ptotal = pgas + pH2O
738 = pgas + 21
Solve for pgas.
Then use PV = nRT and solve for n = number of moles.
Finally, n = grams/molar mass. You know molar mass and n, solve for grams acetylene.
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure
V = Volume
n = moles of gas
R = ideal gas constant
T = temperature in Kelvin
First, we need to convert the temperature from Celsius to Kelvin:
T (Kelvin) = T (Celsius) + 273.15
T (Kelvin) = 23 + 273.15
T (Kelvin) = 296.15 K
Next, we need to calculate the partial pressure of acetylene (C2H2) by subtracting the vapor pressure of water from the total pressure:
Partial Pressure of C2H2 = Total Pressure – Vapor Pressure of Water
Partial Pressure of C2H2 = 738 torr - 21 torr
Partial Pressure of C2H2 = 717 torr
Now, we can rearrange the ideal gas law equation to solve for moles (n):
n = PV / RT
n (moles of C2H2) = (Partial Pressure of C2H2) x (Volume) / (R) x (Temperature in Kelvin)
Using the values:
Partial Pressure of C2H2 = 717 torr
Volume = 523 mL = 0.523 L
R = 0.0821 L·atm/mol·K (Ideal gas constant)
n (moles of C2H2) = (717 torr) x (0.523 L) / (0.0821 L·atm/mol·K) x (296.15 K)
Simplifying the equation:
n (moles of C2H2) = (717) x (0.523) / (0.0821) x (296.15)
n (moles of C2H2) ≈ 9.105
Next, we can calculate the molar mass of acetylene (C2H2):
C (atomic mass) = 12.01 g/mol (Carbon)
H (atomic mass) = 1.01 g/mol (Hydrogen)
Molar mass of C2H2 = 2(C) + 2(H) = 2(12.01) + 2(1.01) = 26.04 g/mol
Now, we can calculate the mass of acetylene collected:
Mass (grams) = moles (C2H2) x molar mass (C2H2)
Mass (grams) ≈ 9.105 moles x 26.04 g/mol
Mass (grams) ≈ 237.29202 grams
Therefore, approximately 237.29 grams of acetylene are collected.
To find the mass of acetylene collected, you need to use the ideal gas law equation:
PV = nRT
Where:
P is the total pressure (adjusted to barometric pressure)
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature in Kelvin
First, convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 23 + 273.15
T(K) = 296.15 K
Next, convert the total pressure from torr to atm:
P(atm) = P(torr) / 760
P(atm) = 738 / 760
P(atm) ≈ 0.9711 atm
Now, we need to calculate the partial pressure of the acetylene (C2H2) using the relationship between the total pressure and the vapor pressure of water:
Partial pressure of C2H2 = Total pressure - Vapor pressure of water
Partial pressure of C2H2 = 0.9711 atm - 21 torr / 760 torr/atm
Partial pressure of C2H2 ≈ 0.9434 atm
Now, we can use the ideal gas law equation to find the number of moles of acetylene (n):
PV = nRT
n = PV / RT
n = (0.9434 atm) * (0.523 L) / (0.0821 L·atm/mol·K * 296.15 K)
n ≈ 0.0198 mol
Finally, we can calculate the mass of acetylene collected using the molar mass of acetylene (C2H2):
Molar mass of C2H2 = 2(atomic mass of C) + 2(atomic mass of H)
Molar mass of C2H2 = 2(12.01 g/mol) + 2(1.01 g/mol)
Molar mass of C2H2 = 26.04 g/mol
Mass of acetylene = moles of acetylene * molar mass of acetylene
Mass of acetylene = 0.0198 mol * 26.04 g/mol
Mass of acetylene ≈ 0.516 g
Therefore, approximately 0.516 grams of acetylene are collected.