Calculate the mass of water produced when 1.57g of butane reacts with excess oxygen

Question

Calculate the mass of water produced when 1.57g of butane reacts with excess oxygen

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Thành Công 4 years 2021-07-12T17:54:53+00:00 1 Answers 6 views 0

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  1. thienthanh
    0
    2021-07-12T17:56:38+00:00
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    Explanation:

    So, first you will want to write the balanced chemical equation for this reaction.

    Butane = C_4H_{10}

    2C_4H_{10}+13O_2=>10H_2O+8CO_2

    ^ This ends up being your balanced chemical equation. Now, you can do the math!

    1.57gC_4H_{10}*\frac{1molC_4H_{10}}{58.12gC_4H_{10}}*\frac{10molH_2O}{2molC_4H_{10}}*\frac{18gH_2O}{1molH_2O}

    After plugging this into a calculator, your final mass of water should be:

    2.43gH2O

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )