If possibly please show work
If 1.00 kg of glucose (5.55 mol) is reacted with 34.0 mol oxygen during the process of cellular respiration, which reactant is limiting? Why does your answer seem logical?
C6H12O6 + 6O2 → 6CO2 + 6H2O + Energy
mols C6H12O6 = 5.55 mols
5.55 mols C6H12O6 will require 6(5.55 = 33.3 mols O2.and you have that much 9at 34) so the reaction will occur as written and glucose will be the limiting reagent.
To determine which reactant is limiting, we need to compare the stoichiometric ratios of the reactants. The balanced chemical equation for cellular respiration is:
C6H12O6 + 6O2 -> 6CO2 + 6H2O
From the equation, we can see that 1 mole of glucose reacts with 6 moles of oxygen.
Given:
Mass of glucose = 1.00 kg
Molar mass of glucose (C6H12O6) = 180.16 g/mol
To convert the mass of glucose to moles, we can use the formula:
Moles = Mass / Molar mass
Moles of glucose = 1000 g / 180.16 g/mol = 5.551 mol
Now, we compare the molar ratio of glucose to oxygen:
Glucose:Oxygen = 5.551 mol : 34.0 mol
Since the ratio between glucose and oxygen is less than 1:6, this means that we do not have enough oxygen to fully react with all the glucose. Therefore, oxygen is the limiting reactant.
The answer is logical because cellular respiration is a process that requires a specific ratio of glucose to oxygen to occur. If there is an excess of glucose but a limited supply of oxygen, the reaction cannot proceed fully and some glucose will be left unreacted.
To determine which reactant is limiting, we need to compare the number of moles of glucose with the number of moles of oxygen in the reaction.
Given:
- Mass of glucose = 1.00 kg
- Molar mass of glucose (C₆H₁₂O₆) = 180.16 g/mol
- Molar mass of oxygen (O₂) = 32.00 g/mol
First, we need to convert the mass of glucose to moles:
moles of glucose = mass of glucose / molar mass of glucose
moles of glucose = (1.00 kg * 1000 g/kg) / 180.16 g/mol
moles of glucose = 5550 g / 180.16 g/mol
moles of glucose = 30.82 mol
So, the number of moles of glucose in the reaction is 30.82 mol.
Given:
- Moles of oxygen = 34.0 mol
Comparing the number of moles of glucose (30.82 mol) with the number of moles of oxygen (34.0 mol), we can determine that the limiting reactant is the reactant that is completely consumed in the reaction.
Since there are fewer moles of glucose (30.82 mol) compared to oxygen (34.0 mol), the limiting reactant is glucose. This means that all of the glucose will be used up in the reaction, while some oxygen will be left over.
This answer seems logical because in a chemical reaction, the reactant that is limited in quantity determines the maximum amount of product that can be formed. In this case, the glucose determines the maximum amount of product that can be produced during cellular respiration.
what is the reaction equation?
If more than 34.0/5.55 = 6.18 moles of O2 are used for each mole of glucose, then the O2 will run out first.