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How many grams of magnesium metal are required to produce 44.8 liters of hydrogen gas at STP according to the chemical equation shown below? (Answer: $\pu{48.6 g}$) $$\ce{Mg(s) + 2 HCl(aq) -> MgCl2(aq) + H2(g)}$$

What I have done:
Amount of hydrogen to be produced= 44.8/22.4 = 2 moles of $\ce{H}$ = 2.02 grams of $\ce{H}$.
Amount of magnesium metal needed $$=\frac{\pu{2.02 g} (\text{of}~\ce{H})\cdot \pu{1 mol} (\text{of}~\ce{H2})\cdot \pu{1 mol} (\text{of}~\ce{Mg})\cdot \pu{24.31 g} (\text{of}~\ce{Mg})}{\pu{2.02 g} (\text{of}~\ce{H2})\cdot \pu{1 mol} (\text{of}~\ce{H2})}$$

I'm not getting the correct answer, can someone please explain how to do this using dimensional analysis? Please just tell me only how to do this, do not show it.

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  • $\begingroup$ You don't have moles of $\ce H$. You have moles of $\ce{H2}$. $\endgroup$
    – Ivan Neretin
    Commented Oct 26, 2015 at 13:00
  • $\begingroup$ The molecular weight of Mg is 24.31 g/mol. Moles multiplied by MW gives the mass (in grams). $\endgroup$
    – LiamH
    Commented Oct 26, 2015 at 13:05
  • $\begingroup$ I have updated your post with some chemistry markup (Your attempt confused me a bit too much, so I skipped that). If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details. $\endgroup$
    – Martin - マーチン ♦
    Commented Oct 26, 2015 at 13:39
  • $\begingroup$ Note that, apparently, the question still uses the old definition of STP with a pressure of $p=1\ \mathrm{atm}$ and a molar volume of an ideal gas of $V_\mathrm m\approx22.4\ \mathrm{l/mol}$. Since 1982, according to IUPAC recommendations, STP corresponds to a temperature of $T=273.15\ \mathrm K$ and a pressure of $p=100\,000\ \mathrm{Pa}$. At this temperature and pressure, the molar volume of an ideal gas actually is $V_\mathrm m=22.710\,947(13)\ \mathrm{l/mol}$. $\endgroup$
    – user7951
    Commented Oct 27, 2015 at 17:08
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    $\begingroup$ It is really simple to solve this problem. You want to have 2 moles of H2. So you'll require twice the amount of Mg required to produce one mole of H2. Since 1 mole of Mg produces 1 mole of H2. 2 moles of Mg will produce 2 moles of H2. We dont need to go to the basics to solve this problem. $\endgroup$
    – Quark
    Commented Oct 28, 2015 at 4:52

1 Answer 1

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It is very helpful when you start with using proper notation, i.e. $V(\ce{H2})=44.8~\mathrm{L}$. Proper notation like $M$ for molecular mass, $m$ for mass, and $n$ for amount of substance will help you follow through your thinking.
When do the calculations, always carry the units with every quantity you use.

It is necessary to know a few principle formulae, for example that molecular mass and mass are connected through the amount of substance: $$M = \frac{m}{n}$$ There are a few more you will need and with their repeated use you will get to know them easily.

When attacking this problem, find out how the quantities relate to each other. In this example it is: If I want to produce one mole of hydrogen gas, how many moles of magnesium do I need?

one, because $n(\ce{Mg})\ce{->}n(\ce{H2})$

A key principle (and asumption) in this exercise is the ideal gas law, $$pV=n\mathcal{R}T.$$ You need it to calculate the amount of substance of hydrogen that should be produced.
(You have indirectly used it by dividing the given value by the molar volume of an ideal gas at STP.)

In this case about two moles of hydrogen gas shall be produced, i.e.
$\displaystyle n(\ce{H2})=\frac{pV(\ce{H2})}{\mathcal{R}T}\approx2~\mathrm{mol}$
(Old definition of STP: $p=1~\mathrm{atm}$ and $T=298.15~\mathrm{K}$)

Now you also know the amount of substance of magnesium you need.

two moles, i.e. $n(\ce{Mg}=2~\mathrm{mol}$

You just have to use the molecular mass of magnesium to find the mass of magnesium you need with the above connection.

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  • $\begingroup$ I'm sorry but how did you get 2 moles of magnesium? Also I'm mostly confused on how to solve this using dimensional analysis, no matter what I do I can't seem to find the correct answer. $\endgroup$
    – user3882522
    Commented Oct 26, 2015 at 18:54
  • $\begingroup$ @user3882522 Can I stress Martin’s points of 1) using proper notations, and 2) properly using units? $\endgroup$
    – Jan
    Commented Oct 28, 2015 at 8:58

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