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How many grams of H2O would be produced if 42.5 grams of C2H3Cl3 were reacted with 11.6 grams of O2?
4C2H4Cl3+7O2-->4CO+6H2O+6Cl2
First we find the limiting reagent.
O2
Next we put the coefficient of the limiting reagent and the product into a ratio.
7:6
Next we divide this by the coefficient of the limiting reagent.
7/7=1 : 6/7=0.857
We then multiply the second number in the ratio by the number of moles in the limiting reagent.
0.857 x 0.362= 0.31
Finally, we multiply this result by the GFW of the product
0.31 x 18= 5.58 grams
That final number is your answer.
4C2H4Cl3+7O2-->4CO+6H2O+6Cl2
First we find the limiting reagent.
O2
Next we put the coefficient of the limiting reagent and the product into a ratio.
7:6
Next we divide this by the coefficient of the limiting reagent.
7/7=1 : 6/7=0.857
We then multiply the second number in the ratio by the number of moles in the limiting reagent.
0.857 x 0.362= 0.31
Finally, we multiply this result by the GFW of the product
0.31 x 18= 5.58 grams
That final number is your answer.