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20 October, 23:32

If you start with 13.4g of c3h8 and 4.35g of o2 determine the limiting reagent

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  1. 21 October, 03:23
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    Balance chemical equation for this reaction is,

    C₃H₈ + 10 O₂ → 3 CO₂ + 4H₂O

    According to this equation 44 g of C₃H₈ reacts with 320 g of O₂ to completely consume.

    Now let's calculate amount of O₂ required to completely consume 13.4 g of C₃H₈,

    44 g C₂H₈ was completely consumed by reacting = 320 g of O₂

    13.4 g of C₂H₈ will require = X g of O₂

    Solving for X,

    X = (13.4 g * 320 g) : 44 g

    X = 97.45 g of O₂ is required to consume 13.4 g of C₂H₈

    Now,

    Let's calculate amount of C₂H₈ required to consume 4.35 g of O₂,

    320 g of O₂ was consumed by reacting = 44 g of C₂H₈

    4.35 g of O₂ will require = X g of C₂H₈

    Solving for X,

    X = (4.35 g * 44 g) : 320

    X = 0.598 g of C₂H₈ is required to consume 4.35 g of O₂

    Result:

    As the given amount of C₂H₈ requires 97.45 g of O₂, but we are provided only with 4.35 g, it means the oxygen will consume first.

    So, Oxygen is the limiting reagent.
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