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17 November, 18:53

How many grams C3H7OH can be made by reacting with 7.3L of CO2 at STP

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  1. 17 November, 21:44
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    6.54g of C3H7OH

    Explanation:

    Step 1:

    Determination of the number of mole of CO2 that occupy 7.3L at stp.

    This can be obtained as follow:

    1 mole of a gas occupy 22.4L at stp.

    Therefore, Xmol of CO2 will occupy 7.3L at stp i. e

    Xmol of CO2 = 7.3/22.4

    Xmol of CO2 = 0.326 mole.

    Therefore, 0.326 mole of CO2 was used in the reaction.

    Step 2:

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

    6CO2 + 8H2O - > 2C3H7OH + 9O2

    Step 3:

    Determination of the number of mole of C3H7OH produced from the reaction. This is illustrated below:

    From the balanced equation above,

    6 moles of CO2 reacted to produce 2 moles of C3H7OH.

    Therefore, 0.326 mole of CO2 will react to produce = (0.326 x 2) / 6 = 0.109 mole of C3H7OH.

    Step 4:

    Conversion of 0.109 mole of C3H7OH to grams. This is illustrated below:

    Number of mole of C3H7OH = 0.109 mole.

    Molar mass of C3H7OH = (12x3) + (7x1) + 16 + 1 = 60g/mol

    Mass of C3H7OH = ... ?

    Mass = mole x molar mass

    Mass of C3H7OH = 0.109 x 60

    Mass of C3H7OH = 6.54g.

    Therefore, 6.54g of C3H7OH is produced from the reaction.
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