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Question : 61 of 180
Marks:
+1 ,
-0
Solution:
Mass of
20 ‌ mL of acetic acid
= 20 × 1 = 20 g Mass of
20 ‌ mL of ethyl alcohol
= 20 × 0.7 = 14 g For the reaction,
+ ‌ CH 3 ‌ COO ‌ C 2 H 5 60 g of acetic acid requires
= 46 g of ethyl alcohol
20 g of acetic acid requires
= ‌ × 20 = 15.3 g of ethy
alcohol
Therefore, ethyl alcohol is the limiting reagent.
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