calculate the volume of oxygen required for cmmplete combution of20cm3 of (C2H2) ethyne
Use Gay-Lussac's Law of Combining Volumes.
complete conbusion of 2cm3 of (C2H2) ethyne
To calculate the volume of oxygen required for the complete combustion of ethyne (C2H2), we need to understand the balanced chemical equation for the combustion reaction.
The balanced chemical equation for the complete combustion of ethyne is:
2C2H2 + 5O2 → 4CO2 + 2H2O
From the balanced equation, we can see that 2 moles of ethyne react with 5 moles of oxygen gas to produce 4 moles of carbon dioxide and 2 moles of water.
To calculate the volume of oxygen gas required, we need to convert the given volume of ethyne to moles using the ideal gas law.
Step 1: Convert the volume of ethyne from cm3 to liters
Given: Volume of ethyne = 20 cm3
1 L = 1000 cm3
=> Volume of ethyne = 20 cm3 ÷ 1000 = 0.02 L
Step 2: Convert the volume of ethyne from liters to moles
We can use the ideal gas law to calculate the number of moles.
PV = nRT
Since we are given the volume and assuming standard temperature and pressure (STP), we can assume P = 1 atm and T = 273 K.
R = 0.0821 L·atm/mol·K
0.02 L * 1 atm = n * 0.0821 L·atm/mol·K * 273 K
=> n = (0.02 L * 1 atm) / (0.0821 L·atm/mol·K * 273 K)
=> n = 0.000956 mol
Step 3: Calculate the volume of oxygen gas required
From the balanced equation, we know that 2 moles of ethyne react with 5 moles of oxygen gas.
Therefore, for 0.000956 mol of ethyne, we need (0.000956 mol * 5 mol) / 2 = 0.00239 mol of oxygen gas.
Now, we can use the ideal gas law again to convert this number of moles to volume.
PV = nRT
Assuming standard temperature and pressure (STP), we can assume P = 1 atm and T = 273 K.
R = 0.0821 L·atm/mol·K
0.00239 mol * V = 1 atm * 0.0821 L·atm/mol·K * 273 K
=> V = (1 atm * 0.0821 L·atm/mol·K * 273 K) / 0.00239 mol
=> V ≈ 95.8 L
Therefore, approximately 95.8 L of oxygen gas is required for the complete combustion of 20 cm3 of ethyne.