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For the balanced equation shown below, if 57.1 grams of K2O were reacted with 14.9 grams of H2O, how many grams of KOH would be produced?
K2O+H2O=>2KOH
57.1/96 = 0.5947916666666667 mol
14.9/18 = 0.8277777777777778 mol
K2O is the limiting reagent.
1 : 2
0.5947916666666667 (2) = 1.189583333333333 mol
1.189583333333333 (57) = 67.80625g
For the balanced equation shown below, if 57.1 grams of K2O were reacted with 14.9 grams of H2O, how many grams of KOH would be produced?
K2O+H2O=>2KOH
57.1/96 = 0.5947916666666667 mol
14.9/18 = 0.8277777777777778 mol
K2O is the limiting reagent.
1 : 2
0.5947916666666667 (2) = 1.189583333333333 mol
1.189583333333333 (57) = 67.80625g