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im an engine, octane combines with oxygen to form carbon dioxide and water. if 22.8 grams of octane combine completly with 80 grams of oxygen to form 70-4 grams of carbon dioxide , what mass of water if formed??

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  1. use the law of conservation of mass.

    Mass in = mass out

    Mass of octane + Mass of O2 = Mass of CO2 + mass of H2O

    22.8+80=70.4+Mass of H2O

    Mass of H2O= 32.4


  2. 2-C8H18+25-O2 ----> 16-CO2+18-H2O

    70.4g(1mol/44g) = 1.6mol(18/16) = 1.8mol H20(18g/1mol) = 32.4g H20

    make it a good day


  3. Step 1:  Know use formulæ, rather than names.  Octane = C₈H₁₈.  Carbon dioxide = CO₂.  Water = H₂O.  And of course, oxygen = O₂.

    Step 2:  Write the basic chemical equation, as stated above: C₈H₁₈ + O₂ → CO₂ + H₂O

    Step 3:  Balance the equation.

    Step 4:  Find the molar weight of each component.

    Step 5:  Use that the find the number of moles of each component.

    Step 6:  Find the limiting factor, if any.  Use that to determine the number of moles of reaction.

    Step 7:  Use that to find the number of moles of each product.

    Step 8:  Use that to find the product weights.

    If you're in Chem 1, things are likely to be simpler.  If you can rely on this, then remember that the goesintas = the comesoutas.  You have 22.8 grams + 80 grams going in, and 70.4 grams + ??? grams coming out.  Use algebra for that:

    .....22.8 + 80 = 70.4 + X

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