Question

Consider the complete combustion of glucose (C6H12O6) with O2 and calculate the moles of CO2 produced when 1.02 g of glucose is reacted with 25 mL of O2 at body temperature (37 ºC) and 0.970 atm.

Answer 1

Answer: The moles of carbon dioxide produced is 0.00095 moles

Step-by-step explanation:

• For Glucose:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Given mass of glucose = 1.02 g

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

`\text{Moles of glucose}=(1.02g)/(180.16g/mol)=0.0057mol`

• For oxygen gas:

To calculate the number of moles, we use ideal gas equation:

`PV=nRT`

where,

P = pressure of the gas = 0.970 atm

V = Volume of the gas = 25 mL = 0.025 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of the gas =
`37^oC=[37+273]K=310K`

R = Gas constant =
`0.0821\text{ L atm }mol^(-1)K^(-1)`

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

`0.970atm* 0.025L=n* 0.0821\text{ L atm }mol^(-1)K^(-1)* 310K\\\\n=(0.970* 0.025)/(0.0821* 310)=0.00095mol`

The chemical equation for the combustion of glucose follows:

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

By Stoichiometry of the reaction:

6 moles of oxygen gas reacts with 1 mole of glucose

So, 0.00095 moles of oxygen gas will react with =
`(1)/(6)* 0.00095=0.00016mol` of glucose

As, given amount of glucose is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

6 moles of oxygen gas produces 6 moles of carbon dioxide

So, 0.00095 moles of oxygen gas will produce =
`(6)/(6)* 0.00095=0.00095mol` of carbon dioxide

Hence, the moles of carbon dioxide produced is 0.00095 moles