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

Step-by-step 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}}=(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 =
`(6)/(1)* 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=(nRT)/(P)=(0.783* 0.0821* 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 =
`(6)/(1)* 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=(nRT)/(P)=(0.783* 0.0821* 310)/(0.960)=20.8L`

Thus volume of
`CO_2` produced is 20.8 Liters