In the reaction between glucose and oxygen,10.0 g of glucose reacts and 7.50 L of carbon dioxide is formed.What is the percent yield if the density of CO₂ is 1.26 g/L?
C₆H₁₂O₆(s) + 6 O₂(g) → 6 CO₂(g) + 6 H₂O(l)

Question

Quizplus · Accepted Answer

First, calculate the molar mass of glucose (C6H12O6) which is 180.16 g/mol. The reaction shows that 1 mole of glucose produces 6 moles of CO2. Given 10.0 g of glucose, this is 10.0 g / 180.16 g/mol = 0.0555 moles of glucose. Theoretically, this should produce 0.0555 moles * 6 = 0.333 moles of CO2. The molar volume of a gas at STP is 22.4 L/mol, but since we're given the density of CO2 as 1.26 g/L, we use the mass of CO2 produced to find the theoretical yield. The molar mass of CO2 is 44.01 g/mol, so 0.333 moles * 44.01 g/mol = 14.65 g of CO2 theoretically. Given the actual production of 7.50 L of CO2 at 1.26 g/L, the actual mass of CO2 produced is 7.50 L * 1.26 g/L = 9.45 g. The percent yield is then (9.45 g / 14.65 g) * 100% = 64.5%.

In the Reaction Between Glucose and Oxygen,10