Question

In the reaction`Mg (s) + 2HCl (aq) -> H_2 (g) + MgCl_2 (aq)`, how many grams of hydrogen gas will be produced from 125.0 milliliters of a 6.0 M HCI in an excess of Mg?

Answer 1

`0.756g`

Explanation:

I am assuming that `6.0 M HCl` is a typo, and that it should be `6.0 molL^-1 HCl`, since that makes sense in the equation.

First we have to find the amount of `HCl` in the solution. We use the formula `n = cV` where `n` is the amount of substance in moles, `c` is the concentration of the solution in moles per litre, and `V` is the volume of the substance in litres.

`n(HCl) = 6.0molL^-1 xx 0.125L = 0.75mol`

Then we find out how many moles of hydrogen gas (`H_2`) are produced. In the formula we see `2HCl`, and `H_2`. This means there is 1 mole of `H_2` for every 2 moles of `HCl` so to find the amount of `H_2` we use:

`n(H_2) = 1/2 xx 0.75mol = 0.375mol`

Now we find the molar mass of the `H_2` molecules, by adding together the atomic weights of the constituent molecules. In this case: `1.008 + 1.008 = 2.016`. Then we use the formula `m = nM` where `m` is the mass of the substance in grams, and `M` is the molar mass of the substance in grams per mole.

`m(H_2) = 0.375mol xx 2.016gmol^-1 = 0.756g`