Question

What weight of magnesium will dissolve in 10 cm^3 of 1 mol/dm^3 HCl and what volume of hydrogen will be produced? Thanks in advance!

Answer 1

0.06 g of `Mg` will dissolve.
112 ml of `H_2` at STP will be produced.

Explanation:

First We will consider the equation of reaction of HCl with Mg.

`Mg + 2HCl rarr MgCl_2 + H_2`

That means, 1 mol of `Mg` reacts with 2 mol of `HCl` to produce 1 mol of `MgCl_2` and 1 mol `H_2`.

The solution contains 1 mol `HCl` in 1 `dm^3`. or, the strength of the solution is 1 `mol L^-1`

So, Amount of `HCl` in 10 `cm^3` or 10 `ml` = 0.01 mol

That means, there is only 0.01 mol `HCl` for reaction.

So, `0.01/2` mol = 0.005 mol Mg will dissolve in the solution and produce 0.005 mol of `H_2` too. [Observe the mole ratio]

`0.005` mol Mg = `12 * 0.005` g = `0.06` g

Before finding the volume of hydrogen, Let's assume the gas is at STP.

1 mol of any gas takes volume of 22.4 L at STP

0.005 mol of any gas will take volume of `22.4 * 0.005` L at STP

                                                              =  0.112 L at STP

                                                              = 112 ml at STP