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

Answer 1

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

Answer 2

the limiting is H2O SINCE IITS THE LEAST

COMES OUT TO BE 28.9 G H2O

Answer 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.

Answer 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

Answer 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.

Answer 6

oh nvmd yes the LR is O2

Answer 7

WHATS THE GRAM OF H2O THEN?

Answer 8

It's 28.9 g.