in eqn, Mg + 2H+ giving Mg++ + H2, what is the minimum value of 4M HCL required to dissolve Mg?

Question

in the above question i have used 2H+ to show a single positive H ion and Mg++ to show dipositive Mg ion.the other + signs are for showing the reaction. in the H2 2 has to be written down but i couldn't do it in the computer. the answer says it is 50 cm cube as 1 mole of Mg reacts with 500 cm cube of 4M HCL. but there is no HCL in the equation, so how is this statement true and how was 50 cm cube derived?

Answer 1

Use the complete balanced equation:

Mg + 2HCl --> Mg+2 + 2Cl-2 + H2

The concentration of the HCl is 4 M. One mole of Mg reacts with 2 moles of HCl. We need to know the moles of Mg.

The density of Magnesium is 1.74 g/cc. 50 cc would contain:

1.74 g/cc * 50 cc = 87.0 grams.

The atomic mass of Magnesium is 24.305 g mole. The number of moles is:

87.0 g / 24.305 g/mole = 3.58 moles

This would require 3.58 * 2 = 7.16 moles of HCl

7.16 moles / 4 moles/Liter = 1.79 liter of 4 M HCl.

Answer 2

Richard's calculations are correct, but I think you've not stated the whole question so he's gone down a different route.

I am guessing that the question was "what is the minimum value of 4M HCl required to dissolve 2.4305 grams of Mg?"

HCl in solution is a mixture of H+ ions and Cl- ions That is where you get from HCl to the given equation.

As you have a 4M solution of HCl, you have *effectively* got a 2M solution of 2H+. So 1 litre of the solution is enough to dissolve 2 moles of Magnesium.

24.305 grams is 1 mole (from the atomic weight)
2.4305 grams is 0.1 mole

2 Moles requires 1000cc
1 Mole requires 500cc
0.1 Mole requires 50cc

Voila!