When 20.0 g C2H6 and 60.0 g of O2 react to form CO2 and H2O, how many grams of water are formed?

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When 20.0 g C2H6 and 60.0 g of O2 react to form CO2 and H2O, how many grams of water are formed?

a. 14.5 g

b. 18.0 g

c. 58.0 g

d. 20.0 g

e. 28.9 g

1 Answer

  • 1. Write the balanced chemical equation for the reaction:

    2C2H6 + 7O2 = 4CO2 + 6H2O

    1. Determine moles C2H6 and O2.

    moles C2H6 = 20.0 grams C2H6 x (1 mole C2H6/30.0692 grams C2H6) = 0.665 moles C2H6

    moles O2 = 60.0 grams O2 x (1 mole O2/32 grams O2) = 1.875 moles O2

    1. Determine limiting reagent. The limiting reagent is the reactant producing the smaller amount of product.

    Is it C2H6?

    Note: 2 moles C2H6 produce 6 moles H2O

    0.665 moles C2H6 x (6 moles H2O/2 moles C2H6) = 1.995 moles H2O

    Is it O2?

    Note: 7 moles O2 produce 6 moles H2O

    1.875 moles O2 x (6 moles H2O/7 moles O2) = 1.607 moles H2O

    Therefore, the LR is O2 since it produces the least amount of product.

    1. Convert moles H2O to grams H2O.

    1.607 mole H2O x (18.02 grams H2O/1 mole H2O) = 28.9 grams H2O

    That’s it!n_n

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