I need help on a stoichiometry

Question

5.    If a student decomposes 58.498 grams of aluminum carbonate and she collects the gas and measures 27.68 grams of carbon dioxide, determine the percent yield for this reaction.           Al₂(CO₃)₃  ----> Al₂O₃  + 3CO₂

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Answer 1

Al₂(CO₃)₃  ----> Al₂O₃  + 3CO₂

moles Al₂(CO₃)₃ used = 58.498 g x 1 mol/233.9898 g = 0.2500 moles

moles CO₂ formed = 0.2500 moles Al₂(CO₃)₃ x 3 moles CO₂/1 mole Al₂(CO₃)₃ = 0.7500 moles CO₂

mass CO₂ (theoretical yield) = 0.7500 moles CO₂ x 44.01 g/mole = 33.01 g CO₂

% yield = actual yield/theoretical yield (x100%) = 27.68 g/33.01 g (x100%) = 83.9%

Answer 2

Greetings,

I'd be more than happy to assist you in solving these types of problems and give you a methodology for others you may encounter. It is important to realize that the collected mass of carbon dioxide is the actual yield from the reaction where the stiochiometry provides the theoretical yield. Feel free to contact me and schedule a session to elaborate on all your needs and help you find success with the rest!