Question

Animals breathe oxygen and release carbon dioxide during cellular respiration according to the equation below. How much oxygen is needed to produce 120 grams of carbon dioxide?

Answer 1

14.544 g of oxygen is needed to produce 120 grams of carbon dioxide.

Step-by-step explanation:

Animals take in oxygen and breathe out carbon dioxide during cellular respiration. The reaction for the metabolism of the food in the animal body is:

`C_6H_(12)O_6 + O_2 \rightarrow 6CO_2 + 6H_2O + Energy`

As can be seen from the reaction stoichiometry that:

6 moles of carbon dioxide gas can be produced from 1 mole of oxygen gas in the process of metabolism of glucose.

Also,

Given :

Mass of carbon dioxide gas = 120 g

Molar mass of carbon dioxide gas = 44 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus, moles of carbon dioxide are:

`moles_(CO_2) = (120 g)/(44 g/mol)`

`moles_(CO_2) = 2.7273 mol`

As mentioned:

6 moles of carbon dioxide gas can be produced from 1 mole of oxygen gas in the process of metabolism of glucose.

1 mole of carbon dioxide gas can be produced from 1/6 mole of oxygen gas in the process of metabolism of glucose.

2.7273 mole of carbon dioxide gas can be produced from
`(1)/(6) * 2.7273` moles of oxygen gas in the process of metabolism of glucose.

Thus, moles of oxygen gas needed = 0.4545 moles

Molar mass of oxygen gas = 32 g/mol

The mass of oxygen gas can be find out by using mole formula as:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Mass\ of\ oxygen\ gas = Moles * Molar mass}`

`Mass\ of\ oxygen\ gas = 0.4545 * 32}`

`Mass\ of\ oxygen\ gas = 14.544 g`

14.544 g of oxygen is needed to produce 120 grams of carbon dioxide.

Answer 2

87.3 g

Step-by-step explanation:

The cellular respiration can be represented through the following equation.

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

We can establish the following relations:

• The molar mass of CO₂ is 44.01 g/mol.
• The molar ratio of CO₂ to O₂ is 6:6.
• The molar mass of O₂ is 32.00 g/mol.

The mass of O₂ that produces 120 g of CO₂ is:

`120gCO_(2).(1molCO_(2))/(44.01gCO_(2)) .(6molO_(2))/(6molCO_(2)) .(32.00gO_(2))/(1molO_(2)) =87.3gO_(2)`