2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2 (g)
How many L of CO2 gas (measured at STP) will be formed during the combustion of 50.0L C2H2 gas
(measured at STP)?
The equation you wrote is not balanced. I've corrected it.
2 C2H2(g) + 5 O2(g) ==> 4 CO2(g) + 2 H2O (g)
50 L C2H2 x (4 L CO2/2 L/2 L C2H2) 50 L* 4L/2L = 100 L @ STP
Note that when gases are involved in the problem and you want volume you can take a shortcut and use L in place of moles. I've done that above. The long way would be
mols C2H2 = 50/22.4 = 2.232
mols CO2 = 2.232 mol C2H2 x (4 mols CO2/2 mol C2H2) = 2.232 x 4/2 = 4.464 moles
Then 4.464 moles x 22.4 L/mol = 100 L @ STP
thank you.
To determine the amount of CO2 gas produced during the combustion of 50.0 L of C2H2 gas, we need to use the stoichiometry of the reaction and the concept of ideal gas laws.
The balanced chemical equation shows that for every 2 moles of C2H2 gas, 4 moles of CO2 gas are produced. We can use this information to set up a conversion ratio:
2 moles of C2H2 gas --> 4 moles of CO2 gas
Now we need to convert the amount of C2H2 gas from liters to moles. To do this, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (at STP = 1 atm)
V = volume (in liters)
n = moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (at STP = 273 K)
Given that the volume of C2H2 gas is 50.0 L, we can calculate the number of moles using the formula:
n = PV/RT
n = (1 atm) * (50.0 L) / (0.0821 L.atm/mol.K * 273 K)
n ≈ 2.15 moles of C2H2 gas
Now we can use the conversion ratio to calculate the amount of CO2 gas produced:
2.15 moles of C2H2 gas --> X moles of CO2 gas
Since the conversion ratio is 2 moles of C2H2 gas to 4 moles of CO2 gas, we can use a proportion to solve for X:
2.15 moles / 2 moles = X moles / 4 moles
Cross-multiplying and solving for X gives:
X = (2.15 moles * 4 moles) / 2 moles
X = 4.3 moles of CO2 gas
Finally, we can convert moles of CO2 gas to volume using the ideal gas law:
n = PV/RT
V = nRT/P
Since we are measuring volume at STP (1 atm and 273 K), we can substitute these values into the equation:
V = (4.3 moles) * (0.0821 L.atm/mol.K) * (273 K) / (1 atm)
V ≈ 9.08 L of CO2 gas
Therefore, approximately 9.08 liters of CO2 gas will be formed during the combustion of 50.0 L of C2H2 gas at STP.