2CH₃CHO+O₂ forms 2CH₃COOH.

If there is 20gm of acetaldehyde and 10gm of O₂.Then which is limiting reagent?

AND BY WHAT EXTENT?

approach is : we have to find which reagent restricts the use up of 100% reagents i.e. the reagent that gives less amount of products on complete use up. 20 gram i.e. 5/11 moles of acetaldehyde and 5/16 moles of o2 is present. from ch3cho 5/11 moles of acetic acid is obtained. from 5/16 moles of o2 5/8 moles of acetic acid is obtained. so we see less amount of acetic acid is obtained from acetaldehyde . there fore this is the limiting reagent. and the other(o2) is in excess . now by how much?5/11 moles of ch3cho reacts with 5/22 moles of o2(2:1 ratio from balnced equation). remaining o2 is 5/16-5/22=15/176 moles or 2.73 gram.please approve if u feel it is properly explained