Question

calculate the volume of oxygen you would need at 1.00 atm and 298 k, Calculate the volume of dry CO2 produced at body temperature (37 ∘C) and 0.960 atm when 23.5 g of glucose is consumed in this reaction.

Answer 1

Answer: a) Volume of $O_2]$ = 19.2 L

b) volume of $CO_2$ = 20.8 Liters

Explanation:

Combustion is a type of chemical reaction in which fuel is reacted with oxygen to form carbon dioxide and water.

$C_6H_{12}O_6+6O_2 \rightarrow 6CO_2+6H_2O$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{23.5g}{180g/mol}=0.131moles$

a) Volume of $O_2$

According to stoichiometry:

1 mole of glucose require = 6 moles of oxygen

Thus 0.131 moles of glucose require =$\frac{6}{1}\times 0.131=0.783$ moles of oxygen

According to the ideal gas equation:'

$PV=nRT$

P = Pressure of the gas = 1.00 atm

V= Volume of the gas = ?

T= Temperature of the gas = 298 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 0.783

$V=\frac{nRT}{P}=\frac{0.783\times 0.0821\times 298}{1.00}=19.2L$

Thus volume of oxygen required is 19.2 Liters

b) Volume of $CO_2$

According to stoichiometry:

1 mole of glucose produce = 6 moles of $CO_2$

Thus 0.131 moles of glucose require =$\frac{6}{1}\times 0.131=0.783$ moles of $CO_2$

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 0.960 atm

V= Volume of the gas = ?

T= Temperature of the gas = $37^0C$=310 K    

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 0.783

$V=\frac{nRT}{P}=\frac{0.783\times 0.0821\times 310}{0.960}=20.8L$

Thus volume of $CO_2$ produced is 20.8 Liters