Determine the type of chemical reaction, find the limiting reagent, determine the amount of product in grams, and the amount in grams of excess reagent.
6.95 moles of glucose C6H12O6(s) is combusted with 7.54 moles of O2 gas (both products!)
C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O
Chemical reaction: combustion
Molar Mass C6H12O6 = 180 amu. 1moleC6H12O6 = 180gC6H12O6
Molar Mass O2 = 32 amu. 1moleO2 = 32gO2
6.95molesC6H12O6 (180gC6H12O6 / 1moleC6H12O6) = 1250gC6H12O6
7.54molesO2 (32gO2 / 1moleO2) = 241gO2
For every 1mole C6H12O6, there are 6molesO2
Limiting reagent: C6H12O6
Product is 1491
When the problem said that the glucose and oxygen were products, does that mean that they are on the right side of the yield sign? Would that change everything? I do not think I am doing this problem right. Could you please tell/show me what I'm doing wrong? I also don't think I addressed the last question. Please help. Thanks!
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