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How many grams of H2O would be produced if 31.5 grams of C2H5NSCl were reacted with 72.8 grams of O2?
C2H5NSCl+5O2-->2CO2+2H2O+2NO+SO3+HCl
First we find the limiting reagent.
C2H5NSCl
Next we put the coefficient of the limiting reagent and the product into a ratio.
1:2
Next we divide this by the coefficient of the limiting reagent.
1/1=1 : 2/1=2
We then multiply the second number in the ratio by the number of moles in the limiting reagent.
2 x 0.286= 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.
C2H5NSCl+5O2-->2CO2+2H2O+2NO+SO3+HCl
First we find the limiting reagent.
C2H5NSCl
Next we put the coefficient of the limiting reagent and the product into a ratio.
1:2
Next we divide this by the coefficient of the limiting reagent.
1/1=1 : 2/1=2
We then multiply the second number in the ratio by the number of moles in the limiting reagent.
2 x 0.286= 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.