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Stoichiometry questiom help?

C6H12O6(s) +6O2(g) -> 6CO2(g) + 6H2O(l)

The human body needs at least 1.03 x 10-2 mol O2 every minute. If all of this oxygen is used for the
cellular respiration reaction that breaks down glucose, how many grams of glucose does the human body
consume each minute?​

Stoichiometry questiom help? C6H12O6(s) +6O2(g) -> 6CO2(g) + 6H2O(l) The human-example-1
User MaxwellN
by
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1 Answer

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Answer:

0.3096g of glucose per minute

Step-by-step explanation:

Given parameters:

Number of moles of O₂ needed per minute = 1.03 x 10⁻²mol

Unknown:

Mass of glucose the body consumes per minute =?

Solution:

The given balanced equation of the reaction is shown below:

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

We have been given the exact amount of all the oxygen gas required for the cellular respiration process per minute. Using this value, we can know how the reaction will proceed successfully.

From the balanced equation:

6 moles of O₂ reacted with 1 mole of C₆H₁₂O₆

therefore 1.03 x 10⁻² of O₂ will react with
(0.0103mole)/(6);

⇒ 0.00172mole of glucose will react per minute.

To find the mass in grams of the glucose that will react, we use the expression below"

Mass of glucose = Number of moles of glucose x molar mass of glucose

Molar mass of C₆H₁₂O₆ = (12 x 6) + (12 x 1) + (16 x 6) = 180g/mol

Mass of glucose = 0.00172mole x 180g/mole = 0.3096g of glucose per minute

User RobS
by
8.3k points
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