# Magnesium reacts with hydrochloric acid according to the following equation: Mg(s) + 2HCl(aq) -> MgCl_2(aq) +H_2(g). What mass of hydrogen will be obtained if 100 cm^3 of 2.00 mol dm^-3 HCl are added to 4.86 g of magnesium?

Jun 24, 2017

Holy dimensional analysis!

0.403g of ${H}_{2}$

See below if you'd like to learn how, so you don't have to rely on us!

#### Explanation:

So you have the balanced equation, that's a good start.

So what we're looking for is the ratio of hydrogen gas produced per the limiting reactant, let's find that first, by getting this ugly DA out of the way.

I think you mean 2.00mol/dm^3, here, an annoying way to put molarity (e.g. concentration) of that acidic solution.

$\frac{2.00 m o l}{{\mathrm{dm}}^{3}} {\left(\frac{{10}^{-} 1 \mathrm{dm}}{c m}\right)}^{3} \left(\frac{c {m}^{3}}{m L}\right) \left(\frac{{10}^{3} m L}{L}\right)$
$\frac{2.00 m o l}{{\mathrm{dm}}^{3}} \left(\frac{{10}^{-} 3 {\mathrm{dm}}^{3}}{c {m}^{3}}\right) \left(\frac{c {m}^{3}}{m L}\right) \left(\frac{{10}^{3} m L}{L}\right)$
$\frac{2.00 m o l}{L} = 2.00 M$

What a waste of time, same value!

Now that we have the molarity, let's find the volume reacted.

$\frac{100 c {m}^{3}}{1} \left(\frac{m L}{c {m}^{3}}\right) = 100 m L$

So, in a more convenient way, we have 100 mL of a 2.00M solution of $H C l$. Let's convert each of the reactants to moles, now:

HCl: $\frac{2.00 m o l}{.100 L} = 20.0 m o l$

Mg(s): $4.86 g \left(\frac{m o l}{24.305 g}\right) = 0.200 m o l$

Clearly, Mg is the limiting reactant , some quick math tells me all we need to fully react that molar amount is 0.400 mol of $H C l$.

$\frac{0.200 m o l M g}{1} \left(\frac{{H}_{2}}{M g}\right) = 0.200 m o l {H}_{2}$

Then, the question asks for the mass, a matter of simply translating the moles of hydrogen gas to its mass in grams (or whatever else if the question specifies).

$\frac{0.200 m o l}{1} \left(\frac{2.016 g}{m o l}\right) = 0.403 g$