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

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:

$\text{Moles of glucose}=\frac{1.02g}{180.16g/mol}=0.0057mol$

• 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:

$0.970atm\times 0.025L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\\\n=\frac{0.970\times 0.025}{0.0821\times 310}=0.00095mol$

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 = $\frac{1}{6}\times 0.00095=0.00016mol$ 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 = $\frac{6}{6}\times 0.00095=0.00095mol$ of carbon dioxide

Hence, the moles of carbon dioxide produced is 0.00095 moles