Same question, different numbers, same solution....
You asked this question already, but with different numbers. It works the same way regardless of what the numbers are.
This is a limiting reactant problem. Compute the amount of product with each reactant. The lesser amount is the theoretical yield, and tells the limiting reactant.
4KO2(s) + 2H2O(l) --> 4KOH(s) + 3O2(g)
18.1g ........ 18.1g ............?g
18.1g KO2 x (1 mol KO2 / 71.1g KO2) x (1 mol KOH / 1 mol KO2) x (56.1g KOH) / 1 mol KOH) = 14.3g KOH
18.1g H2O x (1 mol H2O / 18.0g H2O) x (2 mol KOH / 1 mol H2O) x (56.1g KOH / 1 mol KOH) = 113g KOH
Therefore, 14.3g KOH is produced and KO2 is the limiting reactant.
(18.1 g H2O) / (18.01532 g H2O/mol) = 1.00470 mol H2O
0.25458 mole of KO2 would react completely with 0.25458 x (2/4) = 0.12729 mole of H2O, but there is more H2O present than that, so H2O is in excess and KO2 is the limiting reactant.
(0.25458 mol KO2) x (4 mol KOH / 4 mol KO2) x (56.10564 g KOH/mol) = 14.3 g KOH
Answers & Comments
Same question, different numbers, same solution....
You asked this question already, but with different numbers. It works the same way regardless of what the numbers are.
This is a limiting reactant problem. Compute the amount of product with each reactant. The lesser amount is the theoretical yield, and tells the limiting reactant.
4KO2(s) + 2H2O(l) --> 4KOH(s) + 3O2(g)
18.1g ........ 18.1g ............?g
18.1g KO2 x (1 mol KO2 / 71.1g KO2) x (1 mol KOH / 1 mol KO2) x (56.1g KOH) / 1 mol KOH) = 14.3g KOH
18.1g H2O x (1 mol H2O / 18.0g H2O) x (2 mol KOH / 1 mol H2O) x (56.1g KOH / 1 mol KOH) = 113g KOH
Therefore, 14.3g KOH is produced and KO2 is the limiting reactant.
(18.1 g KO2) / (71.0971 g KO2) = 0.25458 mol KO2
(18.1 g H2O) / (18.01532 g H2O/mol) = 1.00470 mol H2O
0.25458 mole of KO2 would react completely with 0.25458 x (2/4) = 0.12729 mole of H2O, but there is more H2O present than that, so H2O is in excess and KO2 is the limiting reactant.
(0.25458 mol KO2) x (4 mol KOH / 4 mol KO2) x (56.10564 g KOH/mol) = 14.3 g KOH