Calculate the volume of dry co2 produced at body temperature (37 ∘c) and 0.960 atm when 24.0 g of glucose is consumed in this reaction.

The balanced equation for the above reaction is as follows;
C₆H₁₂O₆ + 6O₂ --> 6CO₂ + 6H₂O
molar ratio of glucose to CO₂ is 1:6
the mass of glucose reacted - 24.0 g
Number of glucose moles reacted - 24.0 g /180 g/mol = 0.133 mol
Number of CO₂ formed when 0.133 mol of glucose reacted -0.133 x 6 = 0.798 mol
since CO₂ is a gas we can use the ideal gas law equation to find the volume occupied 
PV = nRT
where P - pressure - 0.960 atm x 101 325 Pa / atm = 97 272 Pa
n - number of moles - 0.798 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 273 + 37 °C = 310 K
substituting the values in the equation
97 272 Pa x V= 0.798 mol x 8.314 Jmol⁻¹K⁻¹ x 310 K
V = 21.1 dm³
volume occupied by CO₂ is 21.1 dm³

The answer is d.potassium