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Limiting reagent, percent yield and percent excess
JEE Main level
Single correct option
Difficulty 2 of 3
mc189
5.4
g
of Al reacts with
18.25
g
of HCl:
2
Al
+
6
HCl
2
AlCl
X
3
+
3
H
X
2
. The volume of
H
X
2
at STP (22.4 L/mol) is (Al = 27, H = 1, Cl = 35.5)
A
6.72
L
B
5.60
L
C
2.24
L
D
3.36
L
Answer: B
n(Al) = 0.20 mol (needs 0.60 mol HCl); n(HCl) = 0.50 mol, so HCl is limiting. H2 = (3/6) x 0.50 = 0.25 mol, volume = 0.25 x 22.4 = 5.60 L.
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