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17 January, 06:39

Propane (C3H8) burns in a combustion reaction. How many grams of C3H8 are needed to produce 80.3 mols CO2?

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  1. 17 January, 07:03
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    1177.88g of C3H8

    Explanation:

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

    C3H8 + 5O2 - > 3CO2 + 4H2O

    Next we shall determine the number of mole of C3H8 required to produce 80.3 moles of CO2. This is illustrated below:

    From the balanced equation above,

    1 mole of C3H8 reacted to produce 3 moles of CO2.

    Therefore, Xmol of C3H8 will react to produce 80.3 moles of CO2 i. e

    Xmol of C3H8 = 80.3/3

    Xmol of C3H8 = 26.77 moles

    Finally, we shall convert 26.77 moles of C3H8 to grams.

    Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol

    Mole of C3H8 = 26.77 moles

    Mass of C3H8 = ... ?

    Mass = mole x molar mass

    Mass of C3H8 = 26.77 x 44

    Mass of C3H8 = 1177.88g

    Therefore, 1177.88g of C3H8 are needed for the reaction
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