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C3H2(g) + 5 O2 (g) → 3 CO2 (g)+ 4H2O(g)
What volume of oxygen is required to completely combust 0.26L of propane (C3Hg) at STP?

Sagot :

Rspill6idn

Answer:

1.3 L O2

Explanation:

I will assume that C3H2 was actually meant to be C3H8, propane.  I'm unaware that C3H2 is even possible.

C3H8(g) + 5 O2 (g) → 3 CO2 (g)+ 4H2O(g)

The equation is balanced with C3H8.

Since the conditions are STP, we can use the molar conversion factor of 22.4 liters/mole (holds any any gas at STP).  [While the combustion of propane is highly unlikely to result in constant STP conditions, it is a lot easier to do the calculations, so I'm all for it.]

The balanced equation tells us that 1 mole of propane will require 5 moles of O2 for complete combustion, as is indicated here.  That means we'll need 5 moles O2/mole C3H8.

Calculate moles propane in 0.26L:  0.26L/(22.4L/mole) = 0.0116 moles of propane.

Multiply that time 5 to determine the moles O2 required for complete combustion:

(0.0116 moles H3H8)*(5 moles O2/mole C3H8) = 0.058 moles O2

Since we are still at STYP (in theory), the volume of O2 will be:

(0.058 moles O2)*(22.4L/mole) = 1.3 L O2 (2 sig figs)

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