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How many grams of SO3 would be produced if 85.7 grams of C4H8S2 were reacted with 297 grams of O2?
C4H8S2+9O2-->4CO2+4H2O+2SO3
First we find the limiting reagent.
C4H8S2
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 0.714= 1.428
Finally, we multiply this result by the GFW of the product
1.428 x 80= 114.24 grams
That final number is your answer.
C4H8S2+9O2-->4CO2+4H2O+2SO3
First we find the limiting reagent.
C4H8S2
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 0.714= 1.428
Finally, we multiply this result by the GFW of the product
1.428 x 80= 114.24 grams
That final number is your answer.