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For the balanced equation shown below, what would be the limiting reagent if 35.6 grams of C4H10S2 were reacted with 51.6 grams of O2?
2C4H10S2+19O2=>8CO2+10H2O+4SO3
2C4H10S2 + 19O2
35.6g 51.6g
122g/mol 32g/mol
.291=mol 1.61=mol
x 19 x 2
5.529 3.22
The limiting reagent is 19O2
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2C4H10S2+19O2=>8CO2+10H2O+4SO3
2C4H10S2 + 19O2
35.6g 51.6g
122g/mol 32g/mol
.291=mol 1.61=mol
x 19 x 2
5.529 3.22
The limiting reagent is 19O2
Don't understand how to do it? Click Here to look at step by step examples!