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. Write the balanced chemical equation for the reaction:
2C2H6 + 7O2 = 4CO2 + 6H2O
- 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
- 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.
- Convert moles H2O to grams H2O.
1.607 mole H2O x (18.02 grams H2O/1 mole H2O) = 28.9 grams H2O