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

1. This is a limiting reagent problem. Do you know how to do standard stoichiometry problems?

2. the limiting is H2O SINCE IITS THE LEAST

COMES OUT TO BE 28.9 G H2O

3. The limiting reagent is not H2O and grams H2O not equal to 28.9. If you will post your work I'll find the error.

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

2C2H6 + 7O2 = 4CO2 + 6H2O

2. 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

3. 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.

3. Convert moles H2O to grams H2O.

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

5. You're right. You don't have an error in this last post. The limiting reagent is O2 (not H2O) and the 28.9 g H2O is correct.

6. oh nvmd yes the LR is O2

7. WHATS THE GRAM OF H2O THEN?

8. It's 28.9 g.

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