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Chemistry help?

Given the following equation:

2C4H10+13O2—>8CO2+10H20

How many grams of the C4H10 are needed to produce 26.00 g of H2O

3 Answers

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  • 3 months ago

    (26.00 g H2O) / (18.01532 g H2O/mol) x (2 mol C4H10 / 10 mol H2O) x

    (58.1222 g C4H10/mol) = 16.78 g C4H10

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  • 3 months ago

    Butane + Oxygen ➜ Carbon Dioxide + water

    2C₄H₁₀ + 13O₂ ➜ 8CO₂ + 10H₂O

    molecular weights

    C = 12

    H = 1

    O = 16

    2C₄H₁₀ = 2•58 = 116

    13O₂ = 13•32 = 416

    8CO₂ = 8•44 = 352

    10H₂O = 10•18 = 180

    check 116+416 = 352+180 = 532

    116 grams of C₄H₁₀ + 416 grams of O₂ ➜ 352 grams of CO₂ + 180 grams of H₂O

    2 mole of C₄H₁₀ + 13 moles of O₂ ➜ 8 mole of CO₂ + 10 moles of H₂O

    How many grams of the C4H10 are needed to produce 26.00 g of H2O

    ratio of C₄H₁₀ to H₂O is 116/180

    116/180 = x/26

    x = 16.8 g

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  • Anonymous
    3 months ago

    26 g of water is about 1.5 moles.  Moles of butane = 1.5 moles H2O * 2 moles butane / 10 moles water.  Multiply moles of butane by its molar mass to get mass.

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