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24 November, 12:07

What mass of O2 was used up in the reaction with an excess of SO2 gas if 14.2 g of sulfur trioxide is formed?

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  1. 24 November, 12:16
    0
    2.84g

    Explanation:

    Step 1:

    The balanced equation for the reaction. This is given below:

    2SO2 + O2 - > 2SO3

    Step 2:

    Determination of the mass of O2 that reacted and the mass of SO3 produced from the balanced equation.

    This is illustrated below:

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

    Mass of O2 from the balanced equation = 1 x 32 = 32g

    Molar mass of SO3 = 32 + (16x3) = 80g/mol

    Mass of SO3 from the balanced equation = 2 x 80 = 160g

    Summary:

    From the balanced equation above,

    32g of O2 reacted to produce 160g of SO3.

    Step 3:

    Determination of the mass of O2 needed to produce 14.2g of SO3.

    This can be achieved as shown below:

    From the balanced equation above,

    32g of O2 reacted to produce 160g of SO3.

    Therefore, Xg of O2 will react to produce 14.2g of SO3 i. e

    Xg of O2 = (32 x 14.2) / 160

    Xg of O2 = 2.84g

    Therefore, 2.84g of O2 is needed to produce 14.2g of SO3.
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