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How many liters of water vapor can be produced if 102 grams of methane gas (CH4) are combusted at 315 K and 1.2 atm? Show all of the work used to solve this problem. CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g)

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transitionstate
Answer:
             V  =  274.77 L

Solution:

First find out the moles of CH₄ as,

                                      Moles  =  Mass / M/mass

                                      Moles  =  102 g / 16 g.mol⁻¹

                                      Moles  =  6.375 mol

Now, according to equation,

                    1 mole CH₄ produced  =  2 moles of H₂O vapours
So,
       6.375 mol of CH₄ will produce  =  X moles of H₂O vapours

Solving for X,
                     X  =  (6.375 mol × 2 mol) ÷ 1 mol

                     X  =  12.75 mol of H₂O 

Now, find out volume of H₂O as,

                                      P V  =  n R T
Or,
                                      V  =  n R T / P

Putting values,

            V  =  (12.75 mol × 0.0821 atm.L.mol⁻¹.K⁻¹ × 315 K) ÷ 1.2 atm

            V  =  274.77 L

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