CaC2 + 2H2O = C2H2 + Ca(OH)2
If 40.0g of calcium carbide (CaC2) reacts with 25.0g of water, how many liters of acetylene (C2H2) can be produced, assuming the reaction goes to completion at 60F and 763 mmHg
convert g to moles in the reaction.
figure how many L per mole of gas at that temp/pressure
To solve this problem, we need to determine the limiting reactant, which is the reactant that will be completely consumed and thus determines the maximum amount of product that can be formed.
1. Start by calculating the number of moles for each reactant. The molar mass of CaC2 is 64.10 g/mol, and the molar mass of H2O is 18.02 g/mol.
Number of moles of CaC2 = mass of CaC2 / molar mass of CaC2
Number of moles of CaC2 = 40.0 g / 64.10 g/mol
Number of moles of H2O = mass of H2O / molar mass of H2O
Number of moles of H2O = 25.0 g / 18.02 g/mol
2. Use the balanced equation to determine the stoichiometry between CaC2 and C2H2. According to the balanced equation, 1 mole of CaC2 produces 1 mole of C2H2.
3. Based on the stoichiometry, compare the number of moles of CaC2 and H2O. The reactant with the smaller number of moles will be the limiting reactant.
4. Calculate the number of moles of C2H2 produced using the limiting reactant (CaC2). According to the stoichiometry, 1 mole of CaC2 produces 1 mole of C2H2.
5. Convert the moles of C2H2 to volume using the ideal gas law. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
6. Convert the volume from liters to another applicable unit if necessary.
Let's go through the calculations:
1. Moles of CaC2:
Number of moles of CaC2 = 40.0 g / 64.10 g/mol ≈ 0.624 mol
2. Moles of H2O:
Number of moles of H2O = 25.0 g / 18.02 g/mol ≈ 1.387 mol
3. Comparing the number of moles:
CaC2: 0.624 mol
H2O: 1.387 mol
Since CaC2 has fewer moles, it is the limiting reactant.
4. Moles of C2H2 (based on CaC2):
Number of moles of C2H2 = 0.624 mol
5. Calculating volume using the ideal gas law:
PV = nRT
We have:
P = 763 mmHg
V (unknown)
n = 0.624 mol
R = 0.0821 L·atm/mol·K
T = 60°F = 15.6°C = 288.75 K
Solving for V:
V = (nRT) / P
V = (0.624 mol)(0.0821 L·atm/mol·K)(288.75 K) / (763 mmHg)
Note: Before plugging values into the equation, make sure to convert temperature to Kelvin and pressure to atmospheres if necessary.
6. Converting from mmHg to liters:
1 atm = 760 mmHg
V = (0.624 mol)(0.0821 L·atm/mol·K)(288.75 K) / (763 mmHg) * (1 atm / 760 mmHg)
Finally, calculate the V:
V ≈ 0.155 L or 155 mL
Therefore, approximately 0.155 liters (or 155 mL) of acetylene (C2H2) can be produced when 40.0g of calcium carbide (CaC2) reacts with 25.0g of water, assuming the reaction goes to completion at 60°F and 763 mmHg.