In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: if 5.97 g of glucose reacts and 1.44 l of co2 gas is collected at 293 k and 0.984 atm, what is the percent yield of the reaction?

The reaction is written below:

C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂

An amount of 5.97 g glucose would have a theoretical yield of:

5.97 g (1 mol glucose/180.16 g)(2 mol CO₂/ 1 mol glucose)(44 g CO₂/1 mol CO₂) = 2.916 g CO₂

% yield = Actual yield/Theoretical yield * 100

PV = nRT
(0.984 atm)(1.44 L CO₂) = n(0.0821 L·atm/mol·K)(293 K)
n = 0.0589 mol
Actual yield = 0.0589 mol * 44 gmol = 2.592 g CO₂

Thus,
% yield = 2.592/2.916 * 100 = 88.9%

Sorry i dont understand the question can you rephase it?