-
This is a limiting reagent problem. Do you know how to do standard stoichiometry problems?
-
the limiting is H2O SINCE IITS THE LEAST
COMES OUT TO BE 28.9 G H2O
-
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.
-
-
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
-
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.
-
oh nvmd yes the LR is O2
-
WHATS THE GRAM OF H2O THEN?
-
-
It's 28.9 g.
