Some Basic Concepts Of Chemistry Question 5
Question: The minimum quantity in gram of $ H _2S $ needed to precipitate 63.5 g of $ C{u^{{2^{+}}}} $ will be $ (C{u^{{2^{+}}}}+H _2S\to Cu _2S+H _2) $ Black
Options:
A) 63.5 g
B) 31.75 g
C) 34 g
D) 2 g
Show Answer
Answer:
Correct Answer: C
Solution:
$ C{u^{2+}}+H _2S\to Cu _2S+H _2 $ 63.5g 1 mole of $ C{u^{2+}} $ requires 1 mole of $ H _2S $ Or 63.5 g of $ C{u^{2+}} $ requires 34 g of $ H _2S $ [Molar mass of $ H _2S $ =2+32=34g] So 1 mole $ H _2S $ is required, i.e, 34 g
