Question

Calculate the volume in liters of CO2 produced at 37 °C and 1.00 atm when 5.60 g glucose (C6H12O6) is used up in the below reaction.

Answer 1

by using pv = nrt ml volume will be find out

and moles equal to given weight upon atomic weight then volume is 0.76 L

Answer 2

Answer:

The volume in liters of CO2 produced at 37 °C is 4.72L.

Explanation:

The reaction will proceed as follows,

$C_6H_1_2O_6(aq)+6O_2(g)\rightarrow 6CO_2(g)+6H_2O(l)$       (1)

Where,

C₆H₁₂O₆=glucose

O₂=Oxygen

CO₂=Carbon dioxide

H₂O=water

From the question we have,

T=37°C +273=310K

Pressure(P)=1 atm

Mass of the glucose(m)=5.60g

The molar mass of the glucose(M)=180g/mol

The number of moles of glucose(n) is given as,

$n=\frac{5.60}{180}=0.031$     (2)

1 mole of glucose produces 6 moles of carbon dioxide. So, 0.031 moles of glucose produces,

$n'=0.031\times 6=0.186$     (3)

From the ideal gas law, we have,

$V=\frac{n'RT}{P}$       (4)

Where,

R=universal gas consatnt=0.082 atm-L/mol-K

By substituting the required values in equation (4) we get;

$V=\frac{0.186\times 0.082\times 310}{1}$

$V=4.72L$

Hence,  the volume in liters of CO2 produced at 37 °C is 4.72L.