Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2
use this reaction to look for a limiting reagent
Mg
n=2.37/24.305=0.098 mol
use the ratio to find the number of moles for H2
1:1
0.098:x
no of moles for H2=0.098mol
use the same procedure
HCl
n=4.56/(1.0079+35.453)=0.125mol
H:HCl
1:2
x:0.125mol
H2 moles =0.0625mol
therefore HCl is a limiting reagent because it produces less number of moles
look fo H2 mass
m=nMr
m=(0.0625)(2.0158)=0.126g
percentage yield =0.11/0.126×100=87.31%
