Question

solve this..

give proper answer ☺️

Answer 1

reaction to be: 
2 CH3CHO + O2 → 2 CH3COOH 

(20 g CH3CHO) / (44.05278 g CH3CHO/mol) = 0.4540 mol CH3CHO 

(10 g O2) / (31.99886 g O2/mol) = 0.3125 mol O2 

0.4540 mole of CH3CHO would react completely with 0.4540 x (1/2) = 0.2270 mole of O2,
but there is more O2 present

than that, so O2 is in excess and

CH3CHO is the limiting reactant. 

(0.4540 mol CH3CHO) x (2 mol CH3COOH / 2 mol CH3CHO) x 
(60.05221 g CH3COOH/mol) = 27 g CH3COOH 

((0.3125 mol O2 initially) - (0.2270 mol O2 reacted)) x (31.99886 g O2/mol) = 
2.7 g O2 left over