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31 October, 08:39

How many grams of H2O can be made from the combustion of 3.75 liters of C7H14 and an excess of O2 at STP?

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  1. 31 October, 09:02
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    21.10g of H2O

    Explanation:

    We'll begin by writing the balanced equation for the reaction. This is given below:

    2C7H14 + 21O2 - > 14CO2 + 14H2O

    From the balanced equation above, 2L of C7H14 produced 14L of H2O.

    Therefore, 3.75L of C7H14 will produce = (3.75 x 14) / 2 = 26.25L of H2O.

    Next, we shall determine the number of mole of H2O that will occupy 26.25L at stp. This is illustrated below:

    1 mole of a gas occupy 22.4L at stp

    Therefore, Xmol of H2O will occupy

    26.25L i. e

    Xmol of H2O = 26.25/22.4

    Xmol of H2O = 1.172 mole

    Therefore, 1.172 mole of H2O is produced from the reaction.

    Next, we shall convert 1.172 mole of H2O to grams. This is illustrated below:

    Number of mole H2O = 1.172 mole

    Molar mass of H2O = (2x1) + 16 = 18g/mol

    Mass of H2O = ... ?

    Mass = mole x molar mass

    Mass of H2O = 1.172 x 18

    Mass of H2O = 21.10g

    Therefore, 21.10g of H2O is produced from the reaction.
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