What volume of oxygen at 25˚C and 725 mm Hg is needed to burn 2.00 L of butane, C4H10 (g), at 725 mm Hg and 25˚C?
2C4H10 + 5O2==> 8CO2 + 10H2O
When dealing with gases with initial and final products at the same temperature and pressure, you can use volume as mols.
Then 2 L C4H10 x (5 mols O2/2 mols C4H10) = ? L O2.
Dr Bob, Check that Oxygen coefficient. => 13 Oxy => 2(13/2)L = 13L Oxy in rxn by your analysis. :-) => 2C4H10 + 13O2 => 4O2 + 5H2O
Another approach is to convert the 2L of Butane to STP volume and divide by 22.4 L/mole. This gives from rxn stoichiometry => 0.086 mole of C4H10 that uses 0.557 mole Oxy. This mole value for Oxy converted to 25C/725mmHg => 0.557mol(22.4L/mole)(298/273)(760/725) = 14.3L Oxy.
Oops! => correction on equation balance... 2C4H10 + 13O2 => 8CO2 + 10H20
Doc, I am positive 13 L is correct. Our difference is in the mols C4H10. If you see something I've done wrong please let me know. Please check your numbers; I didn't go through the 0.557. First, I can't balance an equation. Mine was
2C4H10 + 5O2==> 8CO2 + 10H2O and it should have been
2C4H10 + 13O2 ==> 8CO2 + 10H2O
My statement about taking a shortcut with mols = volume is correct if beginning P and T stay the same so I should have written
2 L C4H10 x (13 mols O2/2 mols C4H10) = 13 L O2 @ 725 mm Hg and 298 K. Now let's go the long way.
With PV = nRT then n = PV/RT and
n = 725 x 2 L/760 x 0.08206 x 298 = 0.078 mols @ STP
0.078 mols C4H10 x (13 mols O2/2 mol C4H10) = 0.507 mols O2 needed @ STP.
Then converting 0.507 mols O2 @ STP to the initial conditions is
V = nRT/P
V = 0.507 x 0.08206 x 298 x 760/725 = 12.996 L = 13 L O2 required..that agrees with the short cut except for some rounding errors.
Dr Bob, I stand corrected… Here’s my rational… After working through the details one should get 13L O₂(g) at specified conditions.
Applying standard volume – in my mind – requires converting to STP conditions then converting back to non-STP conditions. Such still gets the 13L O₂(g).
Vol C₄H₁₀(g) @ STP = 2L C₄H₁₀(g)(725mm/760mm)(273K/298K) = 1.748L C₄H₁₀(g) at STP
Moles C₄H₁₀(g) = 1.748L C₄H₁₀(g)/22.4L/mole = 0.07803 mole C₄H₁₀(g)
Moles O₂(g) required = 13/2(0.07803 mole O₂(g)) =0.507 mole O₂(g)
Vol O₂(g) consumed by 0.07803 mole C₄H₁₀(g) @STP = 0.507 mole O₂(g) x 22.4 L/mole O₂(g) = 11.36 L O₂(g) @ STP
Vol O₂(g) consumed if at 25⁰C & 725mmHg = 11.36L O₂(g) x (760mm/725mm) x (298K/273K) = 13L O₂(g) @ 25⁰C & 725mmHg.
My apologies for being presumptuous. Doc
To determine the volume of oxygen needed to burn 2.00 L of butane, we need to first understand the stoichiometry of the reaction and then apply the ideal gas law.
The balanced equation for the complete combustion of butane is:
2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (g)
From the balanced equation, we can see that 2 moles of butane react with 13 moles of oxygen.
To calculate the volume of oxygen needed, follow these steps:
Step 1: Convert the volume of butane to moles.
Using the ideal gas law equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin, we can rearrange the equation to solve for moles (n).
n1 = (P1 * V1) / (R * T1)
Given:
V1 (volume of butane) = 2.00 L
P1 (pressure of butane) = 725 mm Hg
T1 (temperature of butane) = 25 ˚C = 25 + 273 = 298 K
Substituting the values into the equation:
n1 = (725 mm Hg * 2.00 L) / (0.0821 L·atm/mol·K * 298 K)
n1 = 59.94 mol
Step 2: Calculate the moles of oxygen needed using the stoichiometry of the balanced equation.
From the balanced equation, we know that 2 moles of butane react with 13 moles of oxygen.
n2 = (n1 / 2) * 13
n2 = (59.94 mol / 2) * 13
n2 = 389.61 mol
Step 3: Convert the moles of oxygen to volume.
Using the same ideal gas law equation, we can solve for the volume of oxygen.
V2 = (n2 * R * T2) / P2
Given:
n2 (moles of oxygen) = 389.61 mol
R (gas constant) = 0.0821 L·atm/mol·K
T2 (temperature of oxygen) = 25 ˚C = 25 + 273 = 298 K
P2 (pressure of oxygen) = 725 mm Hg
Substituting the values into the equation:
V2 = (389.61 mol * 0.0821 L·atm/mol·K * 298 K) / 725 mm Hg
V2 = 12.77 L
Therefore, the volume of oxygen needed to burn 2.00 L of butane at 25 ˚C and 725 mm Hg is approximately 12.77 L.