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For the balanced equation shown below, if 48.5 grams of C4H10S were reacted with 91.6 grams of O2, how many grams of H2O would be produced?
C4H10S+8O2=>4CO2+5H2O+SO3
48.5/90 = 0.5388888888888889 mol (8) = 4.311111111111111 mol
91.6/32 = 2.8625 mol
O2 is the limiting reagent.
8 : 5
0.54 (8) = 4.32
18 (5) =
90.1/256 = 0.351953125
0.351953125 (91.6) = 32.23890625g
For the balanced equation shown below, if 48.5 grams of C4H10S were reacted with 91.6 grams of O2, how many grams of H2O would be produced?
C4H10S+8O2=>4CO2+5H2O+SO3
48.5/90 = 0.5388888888888889 mol (8) = 4.311111111111111 mol
91.6/32 = 2.8625 mol
O2 is the limiting reagent.
8 : 5
0.54 (8) = 4.32
18 (5) =
90.1/256 = 0.351953125
0.351953125 (91.6) = 32.23890625g