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How many liters of $\ce{H2(g)}$ at STP is produced per gram of $\ce{Al(s)}$ consumed in the following reaction? $$\ce{2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g)}$$ Express your answer to four significant figures and include the appropriate units.

My attempted solution is included below:

$$ 1.00~\mathrm{g~Al} \left( \dfrac{1~\mathrm{mol~Al}}{26.9815385} \right) \left(\dfrac{3~\mathrm{mol~H_2}}{2~\mathrm{mol~Al}} \right) \left( \dfrac{22.4~\mathrm{L}}{1~\mathrm{mol~H_2}}\right) = 1.245~\mathrm{L}$$

This is being marked as the wrong answer. Where is my error?

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  • $\begingroup$ Seems fine to me. I can only imagine that they expected that you use the standard molar volume constant of perfect gasses, which is slightly higher (and also wrong of course) $V_m=24.5$ L/mol at 298K and $p=p_0$. $\endgroup$
    – Jori
    Commented Oct 24, 2014 at 13:39
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    $\begingroup$ The value $V_\mathrm m=22.4~\mathrm{l/mol}$ for the molar volume of an ideal gas corresponds to the old definition of standard temperature and pressure (STP), i.e. a temperature of $T=273.15~\mathrm K$ and a pressure of $p=1~\mathrm{atm}=101\,325~\mathrm{Pa}$. Since 1982, the recommended STP correspond to a pressure of $p=1~\mathrm{bar}=100\,000~\mathrm{Pa}$. At this pressure, the molar volume actually is $V_\mathrm m=22.710\,947(13)~\mathrm{l/mol}$. However, many textbooks still use the old values. $\endgroup$
    – user7951
    Commented Dec 3, 2016 at 20:31

2 Answers 2

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Your work is correct. I have to assume that this is being graded by a computer, so trivial differences between your answer and the answer that it is expecting matter.

Possible source of error 1: Ideal Gas Law

Instead of using $22.4\ \mathrm{L/mol}$ they may have been expecting you to use the ideal gas law. $$PV = nRT$$ $$101.3\ \mathrm{kPa}\times V = 0.05559\ \mathrm{mol} \times 8.314\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 273.15\ \mathrm K$$ $$V = 1.246\ \mathrm L$$

Possible source of error 2: Units

The computer may be looking for units in liters per gram of aluminum, something like:

$$\frac{\mathrm L}{\mathrm g} \ \text{or}\ \frac{\mathrm L}{\mathrm g\ \ce{Al}}$$

I doubt that this is the case, but it's possible.

Possible source of error 3: Typo

Either you or the person who set up the quiz made a mistake and typed in the answer incorrectly.

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  • $\begingroup$ I just asked my professor and he told me that "the molar volume at STP (1 bar, 0 oC) is slighly not 22.4 L" Looking in the book at can not find the value? $\endgroup$
    – user137452
    Commented Oct 24, 2014 at 18:05
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    $\begingroup$ @user137452 The molar volume at STP ($T=0~\mathrm{^\circ C}=273.15~\mathrm K$, $p=1~\mathrm{bar}=100\,000~\mathrm{Pa}$) is $V_\mathrm m=22.710\,947(13)~\mathrm{l/mol}$ $\endgroup$
    – user7951
    Commented Dec 3, 2016 at 20:45
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The answer is 1.263 One has to use 1 mol gas = 22.414 (at 0 degrees Celsius, 1 atm) = 22.711 (at STP) in the calculation.

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