# 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?

Feb 7, 2016

$0.756 g$

#### Explanation:

I am assuming that $6.0 M H C l$ is a typo, and that it should be $6.0 m o l {L}^{-} 1 H C l$, since that makes sense in the equation.

First we have to find the amount of $H C l$ in the solution. We use the formula $n = c V$ 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.

$n \left(H C l\right) = 6.0 m o l {L}^{-} 1 \times 0.125 L = 0.75 m o l$

Then we find out how many moles of hydrogen gas (${H}_{2}$) are produced. In the formula we see $2 H C l$, and ${H}_{2}$. This means there is 1 mole of ${H}_{2}$ for every 2 moles of $H C l$ so to find the amount of ${H}_{2}$ we use:

$n \left({H}_{2}\right) = \frac{1}{2} \times 0.75 m o l = 0.375 m o l$

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 $m = n M$ where $m$ is the mass of the substance in grams, and $M$ is the molar mass of the substance in grams per mole.

$m \left({H}_{2}\right) = 0.375 m o l \times 2.016 g m o {l}^{-} 1 = 0.756 g$