Ask Question
18 December, 17:34

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?

+5
Answers (1)
  1. 18 December, 20:47
    0
    (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"
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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, ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers