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!
Based on the chemical equation provided, it appears there might be some confusion regarding the reactants and products. Let's clarify some key points to help you solve the problem correctly.
In a chemical equation, the reactants are typically placed on the left side of the arrow (yield sign), while the products are on the right side. In this case, glucose (C6H12O6) and oxygen (O2) should be the reactants because they are being combusted to form the products carbon dioxide (CO2) and water (H2O):
C6H12O6(s) + 6O2(g) -> 6CO2(g) + 6H2O(g)
To determine the limiting reagent, you need to compare the number of moles of each reactant to the stoichiometric ratio given by the balanced chemical equation. The stoichiometric ratio states that 1 mole of C6H12O6 reacts with 6 moles of O2.
Given:
Moles of C6H12O6 = 6.95 mol
Moles of O2 = 7.54 mol
To find the limiting reagent, you can calculate the moles of each reactant needed to react completely based on the stoichiometric ratio:
Moles of C6H12O6 needed = 6.95 mol
Moles of O2 needed = 6.95 mol x (6 mol O2 / 1 mol C6H12O6) = 41.7 mol
Comparing the moles of O2 needed (41.7 mol) with the actual moles of O2 available (7.54 mol), you can see that O2 is the limiting reagent because it is less than the required amount.
Next, let's calculate the amount of product (CO2) that can be formed based on the limiting reagent. From the stoichiometric ratio, we know that for every 1 mole of C6H12O6, 6 moles of CO2 are produced.
Amount of CO2 in grams = moles of C6H12O6 x (6 mol CO2 / 1 mol C6H12O6) x molar mass of CO2
Given:
Molar mass of CO2 = 44 g/mol
Amount of CO2 in grams = 6.95 mol x (6 mol CO2 / 1 mol C6H12O6) x 44 g/mol = 1821.6 g
Therefore, the amount of CO2 product formed is 1821.6 grams.
Finally, to calculate the amount of excess reagent (the reactant that is not completely consumed), you can subtract the moles of the limiting reagent from the moles of the excess reagent. In this case, the excess reagent is C6H12O6, and we already know that we have 6.95 moles of it.
Amount of excess C6H12O6 in grams = moles of C6H12O6 - moles of C6H12O6 used (as the limiting reagent) x molar mass of C6H12O6
Amount of excess C6H12O6 in grams = 6.95 mol - 6.95 mol x molar mass of C6H12O6
Given:
Molar mass of C6H12O6 = 180 g/mol
Amount of excess C6H12O6 in grams = 6.95 mol - 6.95 mol x 180 g/mol = 0 g
Therefore, there is no excess C6H12O6 remaining as the limiting reagent consumes it entirely.
I hope this clarifies the steps and helps you solve the problem correctly!