2B - % Yield calculations |
1) 3.924g of 2-chloropropane reacted with an excess of aqueous sodium hydroxide.
2.955g of propan-2-ol was formed. calculate the % yield of propan-2-ol.
CH3CH2(Cl)CH3 + NaOH à CH3CH2(OH)CH3 + NaCl
3.925g 2.995g
moles = 3.925 moles = 2.995
78.5 61
moles = 0.0500 moles = 0.0490
moles = 0.0500 --------------------> moles = 0.0494
% yield = 0.0490 x 100
0.0500
% yield = 98%
2) Ethanol reacts with ethanoic acid to produce an ester and water.
5.5g of the ester is produced from 4.0g of ethanol and 4.5g of ethanoic acid.
calculate the % yield of the ester.
CH3CH2OH + CH3COOH à CH3COOCH2CH3 + H2O
4.0g 4.5g 5.50g
moles = 4.0 moles = 4.5 moles = 5.50
46 60 88
moles = 0.087 moles = 0.075 moles = 0.0625
(in excess) (limiting reagent)
moles = 0.075 ------> moles = 0.075
% yield = 0.0625 x 100
0.075
% yield = 83.3%