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.
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?
5 answers
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.
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.
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
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
3 answers