Using the following equation 2C2H2 + 5O2 -> 4CO2 + 2H2O
what volume of oxygen would be needed to react with 89.6L of C2H2 at a temp of 273 and a pressure of 101.3 kpa.
I can't seem to figure out my first step. Using PV=NRT should I do initial and final conditions?
I thibk this is a two step question but I don't know what law to use first to find the final volume
on temp, pressue, assume they do not change.
I think you are off track, this is not a gas law problem.
it takes 5 volumes of O2 to react with 2 volumes C2H2.
volumeO2=5/2 * 89.2Liters
Law: You need to read the Law of combining volumes: https://en.wikipedia.org/wiki/%E2%80%93Lussac_law#Law_of_combining_volumes
To solve this problem, you will need to use the ideal gas law equation PV = nRT.
Since you are given the initial volume, temperature, and pressure, you can start by calculating the initial moles of gas using the ideal gas law equation.
Step 1: Calculate the initial moles of C2H2.
To do this, rearrange the ideal gas law equation to solve for n:
n = PV / RT
Substitute the initial values into the equation:
P = 101.3 kPa
V = 89.6 L
R = 8.314 J/(mol·K) (ideal gas constant)
T = 273 K
You need to convert pressure to units of Pascals (Pa) before plugging in the values. 1 kPa = 1000 Pa, so 101.3 kPa = 101300 Pa.
Now, plug the values into the equation:
n1 = (101300 Pa)(89.6 L) / ((8.314 J/(mol·K))(273 K))
Solving this equation will give you the initial moles of C2H2 gas.
Step 2: Use stoichiometry to find the moles of oxygen required.
From the balanced chemical equation:
2C2H2 + 5O2 -> 4CO2 + 2H2O
You can see that 2 moles of C2H2 react with 5 moles of O2. So, you will need to calculate the moles of C2H2, and then use the stoichiometry to find the moles of O2 required.
Step 3: Calculate the final volume of oxygen.
Now that you have the moles of O2 required, you can use the ideal gas law equation again to calculate the final volume of oxygen.
Plug the final values into the ideal gas law equation:
P = 101.3 kPa
V = ? (volume of oxygen)
n = (moles of O2 required from Step 2)
R = 8.314 J/(mol·K) (ideal gas constant)
T = 273 K
Rearrange the equation to solve for V:
V = nRT / P
Substitute the values into the equation and solve for V.
By following these steps, you will be able to find the volume of oxygen needed to react with 89.6 L of C2H2 at a temperature of 273 K and a pressure of 101.3 kPa.
To solve this problem, you can follow these steps:
1. Begin by identifying the given information and the information you need to find. In this case, you are given the volume of C2H2 (89.6L) and need to find the volume of oxygen needed.
2. Convert the initial volume of C2H2 from liters to moles. To do this, you need to divide the volume by the molar volume at STP (Standard Temperature and Pressure), which is approximately 22.4L/mol. So, 89.6L C2H2 / 22.4L/mol = 4 moles of C2H2.
3. Use stoichiometry to determine the mole ratio between C2H2 and O2 from the equation. From the balanced equation, it can be seen that 2 moles of C2H2 react with 5 moles of O2. Therefore, 4 moles of C2H2 will require (4 moles C2H2) x (5 moles O2 / 2 moles C2H2) = 10 moles of O2.
4. Apply the Ideal Gas Law equation, PV = nRT, to find the volume of oxygen. Since you are given the temperature (273K) and pressure (101.3 kPa), you can rearrange the equation to solve for volume.
V = (n * R * T) / P,
where V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.0821 L.atm/mol.K), T is the temperature in Kelvin, and P is the pressure in atmospheres.
5. Plug in the values into the equation:
V = (10 moles * 0.0821 L.atm/mol.K * 273 K) / 101.3 kPa.
Note: Make sure to convert the pressure from kPa to atm by dividing by 101.3 kPa/atm.
6. Perform the calculation:
V = (10 moles * 0.0821 L.atm/mol.K * 273 K) / (101.3 kPa / 101.3 kPa/atm) = 2.08 L.
Therefore, the volume of oxygen needed to react with 89.6L of C2H2 is approximately 2.08 liters.