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24 April, 04:20

When hydrocarbons are burned in a limited amount of air, both CO and CO2 form. When 0.430 g of a particular hydrocarbon was burned in air, 0.446 g of CO, 0.700 g of CO2, and 0.430 g of H2O were formed.

Required:

a. What is the empirical formula of the compound?

b. How many grams of O2 were used in the reaction?

c. How many grams would have been required for complete combustion?

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  1. 24 April, 07:33
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    (a) The empirical formula of the compound is

    m (CxHy) + m (O2) = m (CO) + m (CO2) + m (H2O).

    (b) The grams of O2 that were used in the reaction is 1.146 g

    (c) The amount of O2 that would have been required for complete combustion is 1.401 g.

    Explanation:

    a. m (CxHy) + m (O2) = m (CO) + m (CO2) + m (H2O)

    (b) Using law of conservation of mass from above

    m (O2) = m (CO) + m (CO2) + m (H2O) - m (CxHy)

    m (O2) = 0.446 + 0.700 + 0.430 - 0.430

    m (O2) = 1.146 g

    The grams of O2 that were used in the reaction is 1.146 g

    (c) for complete combustion, we need to oxidized CO to CO2

    Then, 2CO + O2 = 2CO2

    m (add) (O2) = M (O2) * ¢ (O2) / 2 = M (O2) * { (m (CO)) / (2M (CO)) }

    m (add) (O2) = 32 * { (0.446) / (2*28) } = 0.255 g

    Note; Molar mass of O2 = 32, CO = 28

    m (total) (O2) = m (O2) + m (add) (O2)

    m (total) (O2) = 1.146 + 0.255 = 1.401 g

    The amount of that grams would have been required for complete combustion is 1.401 g.

    Note (add) and (total) were used subscript to "m"
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