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How many grams of H2O would be produced if 78.8 grams of C2H3Cl3 were reacted with 34.1 grams of O2?
4C2H4Cl3+11O2-->8CO2+6H2O+6Cl2
First we find the limiting reagent.
O2
Next we put the coefficient of the limiting reagent and the product into a ratio.
11:6
Next we divide this by the coefficient of the limiting reagent.
11/11=1 : 6/11=0.54
We then multiply the second number in the ratio by the number of moles in the limiting reagent.
0.54 x 1.06= 0.572
Finally, we multiply this result by the GFW of the product
0.572 x 18= 10.296 grams
That final number is your answer.
4C2H4Cl3+11O2-->8CO2+6H2O+6Cl2
First we find the limiting reagent.
O2
Next we put the coefficient of the limiting reagent and the product into a ratio.
11:6
Next we divide this by the coefficient of the limiting reagent.
11/11=1 : 6/11=0.54
We then multiply the second number in the ratio by the number of moles in the limiting reagent.
0.54 x 1.06= 0.572
Finally, we multiply this result by the GFW of the product
0.572 x 18= 10.296 grams
That final number is your answer.