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The mass of iron that remained at the end of the experiment : 6.65 g
Given
V HCl gas = 300 cm³
mass Fe = 7 g(Ar Fe = 56 g/mol)
Reaction
Fe(s) + 2HCl(g) → FeCl2(s) + H2(g)
Required
The mass of iron remained
Solution
molar volume for 1 mol gas = 24000 cm³=24 L, so mol HCl :
[tex]\tt \dfrac{300}{24000}=0.0125~mole[/tex]
mol ratio Fe : HCl from equation : 1 : 2, so mol Fe =
[tex]\tt \dfrac{1}{2}\times 0.0125=0.00625[/tex]
mass of Iron(reacted) :
[tex]\tt 0.00625\times 56=0.35~g[/tex]
Mass remained
7 g - 0.35 = 6.65 g