how many mol Fe2 + ions are formed when excess of iron is treated with 50 ml of 4.0 M HCL under inert atmosphere assume no change in volume
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The equation for this reaction is as follows:
2HCl + Fe ---> Fe2+ + H2 + 2Cl-
which can also be written as
2H+ + Fe ---> Fe2+ + H2
The Cl- ions do not take place in the reaction itself so they can be removed.
Thus, simple equation would be:
2HCl + Fe ---> Fe2+ + H2 + Cl-
Now, Molarity x Volume (in litres)
= moles 4M x (.05L)
= 0.2mol (HCl) 0.2 mol HCL x [1 mol Fe2+ / 2mol HCl]
= 0.2/2 = 0.1 mol Fe2+ ions
2HCl + Fe ---> Fe2+ + H2 + 2Cl-
which can also be written as
2H+ + Fe ---> Fe2+ + H2
The Cl- ions do not take place in the reaction itself so they can be removed.
Thus, simple equation would be:
2HCl + Fe ---> Fe2+ + H2 + Cl-
Now, Molarity x Volume (in litres)
= moles 4M x (.05L)
= 0.2mol (HCl) 0.2 mol HCL x [1 mol Fe2+ / 2mol HCl]
= 0.2/2 = 0.1 mol Fe2+ ions
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