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Please Help Me Figure Out This Chemistry Problem.?

Butane gas (C4H10) is commonly used as a portable fuel for stoves,
theoretically, how much oxygen is needed to burn 29kg of butane?
I need an explanation and complete solution.

This entry was posted on Thursday, November 5th, 2009 at 4:25 pm and is filed under Stoves. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Please Help Me Figure Out This Chemistry Problem.?”

1. Ms. Worth on November 5th, 2009 at 9:25 pm

   Here’s the balanced equation.
   2 C₄H₁₀ + 13 O₂ ⇒ 8 CO₂ + 10 H₂O
   One mole of butane = 58 g and one mole of O₂ = 32 g.
   The equation says that 116 g ( 2 moles ) of butane needs 416 g ( 13 moles ) of O₂.
   The problem calls for — not two moles but — one half a mole weight (one quarter the amount in the equation), so it would use one fourth the amount of O₂.
   That would be 104 kg of O₂.
   ________________________________
   Hi, Telle ~
   Is this what you need?

2. tff on November 5th, 2009 at 9:37 pm

   2C4H10+1302=8CO2+10H20
   no of mole of C4H10
   =29/((4*12)+(10*1))
   =29/58
   =0.5 mol
   2 mol of butane needs 10 mol of oxygen
   0.5 mol of butane needs 2.5 mol of oxygen

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