# For the oxidation of glucose (C6H12O6 + 602 + 6H20 + 6CO2), how many moles of oxygen will it take to consume 6 moles of glucose?

The moles of oxygen required to consume 6 moles of glucose has been 36 moles.

The balanced chemical equation for the oxidation of glucose has been:

According to the balanced chemical equation, for oxidation of 1 mole of glucose, 6 moles of oxygen has been required.

Computation for the moles of oxygen

The available moles of glucose has been 6.

The moles of oxygen required has been:

Thus, the moles of oxygen required to consume 6 moles of glucose has been 36 moles.

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Explanation:

Do stoichiometry between glucose and oxygen using the mole to mole ration found in the balanced equation:

mole to mole ratio:

1 mole C6H12O6 : 6 moles O2

6 moles C6H12O6 X 6 moles O2 / 1 mole C6H12O6

= 36 moles O2

## Related Questions

An intermediate step in the process that ultimately produces sulfuric acid, is the catalytic oxidation of SO2 to SO3. Vanadium(V) oxide is the catalyst. The three chemical species in the reaction are in equilibrium as follows: 2SO2(g)+O2(g)⇌2SO3(g) Which of the following are true statements about this reaction? When SO2 is mixed with O2 in a container, the initial rate of the forward reaction (production of SO3) is faster than the initial rate of the reverse reaction (production of SO2). As SO2 is used up and SO3 accumulates, the rate of the forward reaction increases and the rate of the reverse reaction decreases. At equilibrium the rate of production of SO3 reaches zero. At equilibrium the concentration of SO2 will no longer be changing.

When SO₂ is mixed with O₂ in a container, the initial rate of the forward reaction (production of SO₃) is faster than the initial rate of the reverse reaction (production of SO₂).

At equilibrium the concentration of SO₂ will no longer be changing.

Explanation:

Let's consider the following reaction.

2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g)

When SO₂ is mixed with O₂ in a container, the initial rate of the forward reaction (production of SO₃) is faster than the initial rate of the reverse reaction (production of SO₂). TRUE. Initially, when there is no SO₃ in the container, the rate of production of SO₂ is zero.

As SO₂ is used up and SO₃ accumulates, the rate of the forward reaction increases and the rate of the reverse reaction decreases. FALSE. As the equilibrium is reached, the rate of the forward reaction decreases and the rate of the reverse reaction increases.

At equilibrium the rate of production of SO₃ reaches zero. FALSE. At equilibrium, the rate of production of SO₃ is equal to the rate of production of SO₂, which is why the concentrations remain constant.

At equilibrium the concentration of SO₂ will no longer be changing. TRUE. At equilibrium, the concentrations of reactants and products remain constant.

some components of ink are minimally attracted to the stationary phase and very soluble in the solvent. where are these components locatedon the filter paper during chromatography?

Explanation:

Chromatography is defined as technique for separating a mixture of chemicals which can be present in a liquid or gaseous phase.

These chemical components are separated with the help of a stationary medium on which these components travel at a different rates.

The stationary phase could be a TLC plate or silica gel. And, the mobile phase includes the solvent.

For example, in the given situation ink is our mobile phase. Since, some of the components of ink are soluble into the solvent.

Therefore, ink components will travel will the solvent to the top.

Thus, we can conclude that at the top these components located on the filter paper during chromatography.

They’ll have moved the farthest, since the solvent is best at carrying those kinds of materials.

What is the mass in grams of 13.2 amu?