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The answer was: 129.5 grams
For the balanced equation shown below, if 95.4 grams of C6H4Cl2 were reacted with 102 grams of O2, how many grams of CO2 would be produced?
2C6H4Cl2+13O2=>12CO2+2H2O+4HCl
95.4/111.5 = 0.8556053811659193 mol (13) = 11.12286995515695 mol
102/32 = 3.1875 mol (2) = 6.375 mol
O2 is the limiting reagent.
13 : 12
32 (13) = 416
44 (12) = 528
528 / 416 = 1.269230769230769 (102) = 129.4615384615385g
The answer was: 129.5 grams
For the balanced equation shown below, if 95.4 grams of C6H4Cl2 were reacted with 102 grams of O2, how many grams of CO2 would be produced?
2C6H4Cl2+13O2=>12CO2+2H2O+4HCl
95.4/111.5 = 0.8556053811659193 mol (13) = 11.12286995515695 mol
102/32 = 3.1875 mol (2) = 6.375 mol
O2 is the limiting reagent.
13 : 12
32 (13) = 416
44 (12) = 528
528 / 416 = 1.269230769230769 (102) = 129.4615384615385g