Question

determine the volume of carbon dioxide produced at 35°C and 1.13 atm when 8.34 g of glucose was burned after a strenuous exercise. Refer to the following balanced chemical equation

Answer 1

С₆H₁₂O₆ + 6O₂ --> 6CO₂ + 6H₂O;

Volume of carbon dioxide at normal conditions:

V₁(CO₂)=m(glucose)*V_M/Mr(glucose) = 8.34*6*22.4/180 = 6.23 L;

Volume of carbon dioxide at given conditions:

V₂(CO₂)=V₁p₁T₂/p₂T₁=6.23*1*308/(1.131*293)=5.8 L