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22 December, 16:05

The combustion of propane may be described by the chemical equation C 3 H 8 (g) + 5 O 2 (g) ⟶ 3 CO 2 (g) + 4 H 2 O (g) How many grams of O 2 (g) are needed to completely burn 80.9 g C 3 H 8 (g) ?

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  1. 22 December, 16:11
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    294.2g

    Explanation:

    C3H8 + 5O2 - > 3CO2 + 4H2O

    Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

    Molar Mass of O2 = 16x2 = 32g/mol

    Mass of O2 from the balanced equation = 5 x 32 = 160g

    From the equation,

    44g of C3H8 was completely consumed by 160g of O2.

    Therefore, 80.9g of C3H8 will be consume by = (80.9x160) / 44 = 294.2g of O2

    Therefore, 294.2g of O2 is needed
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