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Why are only 6 water molecules are formed in the aerobic degradation of glucose?

I am studying the aerobic degradation of glucose and it seems that for every glucose molecule we should obtain $\ce{10H2O}$ molecules. However, it is known that we only obtain 6.

$\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}$

(I am not going to focus in all products and reactants, but just in the important ones for the formation of water molecules)

First, in the glycolysis, for each molecule of glucose we obtain two water molecules, $\ce{2NADH+}$ and 2 pyruvate molecules. By the oxidation of two pyruvate molecules we obtain $\ce{2NADH+}$ and 2 acetyl Co-A molecules. So we are going to pass twice through the Krebs cycle, obtaining $\ce{6NADH+}$ and $\ce{2FADH2}$, and requiring 4 water molecules.

So, when we arrive to the electron transport chain, we have a negative balance of 2 water molecules, and we have $\ce{10NADH+}$ and $\ce{2FADH2}$. We have been told that for every of these molecules 2 electrons go to the electron transport chain, that means that a total of 24 electrons go to the system. The problem comes here:

$\ce{4e- + 4H+ + O2 = 2H2O}$

So, bearing in mind that we have $24e^-$, 12 water molecules should be formed, so at the end, we have gained 10 water molecules, but we know that the number of water molecules formed should be 6. So, clearly there is something wrong in my explanation. I would be very pleased if you could tell me what is wrong.

Thanks in advance.

Why only 6 water molecules are formed in the aerobic degradation of glucose?

I am studying the aerobic degradation of glucose and it seems that for every glucose molecule we should obtain $\ce{10H2O}$ molecules. However, it is known that we only obtain 6.

$\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}$

(I am not going to focus in all products and reactants, but just in the important ones for the formation of water molecules)

First, in the glycolysis, for each molecule of glucose we obtain two water molecules, $\ce{2NADH+}$ and 2 pyruvate molecules. By the oxidation of two pyruvate molecules we obtain $\ce{2NADH+}$ and 2 acetyl Co-A molecules. So we are going to pass twice through the Krebs cycle, obtaining $\ce{6NADH+}$ and $\ce{2FADH2}$, and requiring 4 water molecules.

So, when we arrive to the electron transport chain, we have a negative balance of 2 water molecules, and we have $\ce{10NADH+}$ and $\ce{2FADH2}$. We have been told that for every of these molecules 2 electrons go to the electron transport chain, that means that a total of 24 electrons go to the system. The problem comes here:

$\ce{4e- + 4H+ + O2 = 2H2O}$

So, bearing in mind that we have $24e^-$, 12 water molecules should be formed, so at the end, we have gained 10 water molecules, but we know that the number of water molecules formed should be 6. So, clearly there is something wrong in my explanation. I would be very pleased if you could tell me what is wrong.

Thanks in advance.

Why are only 6 water molecules formed in the aerobic degradation of glucose?

I am studying the aerobic degradation of glucose and it seems that for every glucose molecule we should obtain $\ce{10H2O}$ molecules. However, it is known that we only obtain 6.

$\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}$

(I am not going to focus in all products and reactants, but just in the important ones for the formation of water molecules)

First, in the glycolysis, for each molecule of glucose we obtain two water molecules, $\ce{2NADH+}$ and 2 pyruvate molecules. By the oxidation of two pyruvate molecules we obtain $\ce{2NADH+}$ and 2 acetyl Co-A molecules. So we are going to pass twice through the Krebs cycle, obtaining $\ce{6NADH+}$ and $\ce{2FADH2}$, and requiring 4 water molecules.

So, when we arrive to the electron transport chain, we have a negative balance of 2 water molecules, and we have $\ce{10NADH+}$ and $\ce{2FADH2}$. We have been told that for every of these molecules 2 electrons go to the electron transport chain, that means that a total of 24 electrons go to the system. The problem comes here:

$\ce{4e- + 4H+ + O2 = 2H2O}$

So, bearing in mind that we have $24e^-$, 12 water molecules should be formed, so at the end, we have gained 10 water molecules, but we know that the number of water molecules formed should be 6. So, clearly there is something wrong in my explanation. I would be very pleased if you could tell me what is wrong.

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canadianer
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I am studying the aerobic degradation of glucose and it seems that for every glucose molecule we should obtain $10H_2O$$\ce{10H2O}$ molecules. However, it is known that we only obtain 6.

$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6 H_2O$$\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}$

(I am not going to focus in all products and reactants, but just in the important ones for the formation of water molecules)

First, in the glycolysis, for each molecule of glucose we obtain two water molecules, $ 2NADH^+$$\ce{2NADH+}$ and 2 pyruvate molecules. By the oxidation of two pyruvate molecules we obtain $2NADH^+$$\ce{2NADH+}$ and 2 acetyl Co-A molecules. So we are going to pass twice forthrough the krebsKrebs cycle, obtaining $6NADH^+$$\ce{6NADH+}$ and $2FADH_2$$\ce{2FADH2}$, and requiring 4 water molecules.

So, when we arrive to the electron transport chain, we have a negative balance of 2 water molecules, and we have $10NADH^+$$\ce{10NADH+}$ and $2FADH_2$$\ce{2FADH2}$. We have been told that for every of these molecules 2 electrons go to the electron transport chain, that means that a total of 24 electrons go to the system. The problem comes here:

$4e^- + 4H^+ + O_2 = 2H_2O$$\ce{4e- + 4H+ + O2 = 2H2O}$

So, bearing in mind that we have $24e^-$, 12 water molecules should be formed, so at the end, we have gained 10 water molecules, but we know that the number of water molecules formed should be 6. So, clearly there is something wrong in my explanation. I would be very pleased if you could tell me what is wrong.

Thanks in advance.

I am studying the aerobic degradation of glucose and it seems that for every glucose molecule we should obtain $10H_2O$ molecules. However, it is known that we only obtain 6.

$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6 H_2O$

(I am not going to focus in all products and reactants, but just in the important ones for the formation of water molecules)

First, in the glycolysis, for each molecule of glucose we obtain two water molecules, $ 2NADH^+$ and 2 pyruvate molecules. By the oxidation of two pyruvate molecules we obtain $2NADH^+$ and 2 acetyl Co-A molecules. So we are going to pass twice for the krebs cycle, obtaining $6NADH^+$ and $2FADH_2$ and requiring 4 water molecules.

So, when we arrive to the electron transport chain, we have a negative balance of 2 water molecules, and we have $10NADH^+$ and $2FADH_2$. We have been told that for every of these molecules 2 electrons go to the electron transport chain, that means that a total of 24 electrons go to the system. The problem comes here:

$4e^- + 4H^+ + O_2 = 2H_2O$

So, bearing in mind that we have $24e^-$, 12 water molecules should be formed, so at the end, we have gained 10 water molecules, but we know that the number of water molecules formed should be 6. So, clearly there is something wrong in my explanation. I would be very pleased if you could tell me what is wrong.

Thanks in advance.

I am studying the aerobic degradation of glucose and it seems that for every glucose molecule we should obtain $\ce{10H2O}$ molecules. However, it is known that we only obtain 6.

$\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}$

(I am not going to focus in all products and reactants, but just in the important ones for the formation of water molecules)

First, in the glycolysis, for each molecule of glucose we obtain two water molecules, $\ce{2NADH+}$ and 2 pyruvate molecules. By the oxidation of two pyruvate molecules we obtain $\ce{2NADH+}$ and 2 acetyl Co-A molecules. So we are going to pass twice through the Krebs cycle, obtaining $\ce{6NADH+}$ and $\ce{2FADH2}$, and requiring 4 water molecules.

So, when we arrive to the electron transport chain, we have a negative balance of 2 water molecules, and we have $\ce{10NADH+}$ and $\ce{2FADH2}$. We have been told that for every of these molecules 2 electrons go to the electron transport chain, that means that a total of 24 electrons go to the system. The problem comes here:

$\ce{4e- + 4H+ + O2 = 2H2O}$

So, bearing in mind that we have $24e^-$, 12 water molecules should be formed, so at the end, we have gained 10 water molecules, but we know that the number of water molecules formed should be 6. So, clearly there is something wrong in my explanation. I would be very pleased if you could tell me what is wrong.

Thanks in advance.

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maxbp
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I am studying the aerobic degradation of glucose and it seems that for every glucose molecule we should obtain $12H_2O$$10H_2O$ molecules. However, it is known that we only obtain 6.

$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6 H_2O$

(I am not going to focus in all products and reactants, but just in the important ones for the formation of water molecules)

First, in the glycolysis, for each molecule of glucose we obtain two water molecules, $ 2NADH^+$ and 2 pyruvate molecules. By the oxidation of two pyruvate molecules we obtain $2NADH^+$ and 2 acetyl Co-A molecules. So we are going to pass twice for the krebs cycle, obtaining $6NADH^+$ and $2FADH_2$ and requiring 24 water molecules.

So, when we arrive to the electron transport chain, we have neither won nor lost anya negative balance of 2 water moleculemolecules, and we have $10NADH^+$ and $2FADH_2$. We have been told that for every of these molecules 2 electrons go to the electron transport chain, that means that a total of 24 electrons go to the system. The problem comes here:

$4e^- + 4H^+ + O_2 = 2H_2O$

So, bearing in mind that we have $24e^-$, 12 water molecules should be formed, so at the end, we have gained 10 water molecules, but we know that the number of water molecules formed should be 6. So, clearly there is something wrong in my explanation. I would be very pleased if you could tell me what is wrong.

Thanks in advance.

I am studying the aerobic degradation of glucose and it seems that for every glucose molecule we should obtain $12H_2O$ molecules. However, it is known that we only obtain 6.

$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6 H_2O$

(I am not going to focus in all products and reactants, but just in the important ones for the formation of water molecules)

First, in the glycolysis, for each molecule of glucose we obtain two water molecules, $ 2NADH^+$ and 2 pyruvate molecules. By the oxidation of two pyruvate molecules we obtain $2NADH^+$ and 2 acetyl Co-A molecules. So we are going to pass twice for the krebs cycle, obtaining $6NADH^+$ and $2FADH_2$ and requiring 2 water molecules.

So, when we arrive to the electron transport chain, we have neither won nor lost any water molecule, and we have $10NADH^+$ and $2FADH_2$. We have been told that for every of these molecules 2 electrons go to the electron transport chain, that means that a total of 24 electrons go to the system. The problem comes here:

$4e^- + 4H^+ + O_2 = 2H_2O$

So, bearing in mind that we have $24e^-$, 12 water molecules should be formed, but we know that the number of water molecules formed should be 6. So, clearly there is something wrong in my explanation. I would be very pleased if you could tell me what is wrong.

Thanks in advance.

I am studying the aerobic degradation of glucose and it seems that for every glucose molecule we should obtain $10H_2O$ molecules. However, it is known that we only obtain 6.

$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6 H_2O$

(I am not going to focus in all products and reactants, but just in the important ones for the formation of water molecules)

First, in the glycolysis, for each molecule of glucose we obtain two water molecules, $ 2NADH^+$ and 2 pyruvate molecules. By the oxidation of two pyruvate molecules we obtain $2NADH^+$ and 2 acetyl Co-A molecules. So we are going to pass twice for the krebs cycle, obtaining $6NADH^+$ and $2FADH_2$ and requiring 4 water molecules.

So, when we arrive to the electron transport chain, we have a negative balance of 2 water molecules, and we have $10NADH^+$ and $2FADH_2$. We have been told that for every of these molecules 2 electrons go to the electron transport chain, that means that a total of 24 electrons go to the system. The problem comes here:

$4e^- + 4H^+ + O_2 = 2H_2O$

So, bearing in mind that we have $24e^-$, 12 water molecules should be formed, so at the end, we have gained 10 water molecules, but we know that the number of water molecules formed should be 6. So, clearly there is something wrong in my explanation. I would be very pleased if you could tell me what is wrong.

Thanks in advance.

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