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For the balanced equation shown below, if 12.3 grams of CH3COF were reacted with 2.37 grams of H2O, how many grams of CH3COOH would be produced?
CH3COF+H2O=>CH3COOH+HF
12.3/62 = 0.1983870967741935 mol
2.37/18 = 0.1316666666666667 mol
H2O is the limiting reagent.
1 : 1
0.1316666666666667 (60) = 7.900000000000002g
For the balanced equation shown below, if 12.3 grams of CH3COF were reacted with 2.37 grams of H2O, how many grams of CH3COOH would be produced?
CH3COF+H2O=>CH3COOH+HF
12.3/62 = 0.1983870967741935 mol
2.37/18 = 0.1316666666666667 mol
H2O is the limiting reagent.
1 : 1
0.1316666666666667 (60) = 7.900000000000002g