Question

What volume of hydrogen gas is produced when `"6 g"` of magnesium reacts?

Answer 1

Well, I got a more exact answer... but you have only allowed yourself one significant figure. Sorry to say, I can only give your answer to zero decimal places.

`V_(H_2(g)) ~~ "6 L"`...

---

I assume you mean `25^@ "C"` and `"1 atm"`, but you know your room temperature better than anyone around you. As for the reaction, I can only assume what it is... Is it not this one?

`"Mg"(s) + 2"HCl"(aq) -> "MgCl"_2(aq) + "H"_2(g)`

If we are on the same page, then for every `"1 mol"` of `"Mg"` consumed, `"1 mol"` of `"H"_2(g)` is made. This was found from the stoichiometry of the reaction, i.e. `color(red)1 xx "Mg"(s) harr color(red)1 xx "H"_2(g)`.

This means that from the molar mass of `"Mg"`:

`6 cancel"g Mg" xx cancel"1 mol Mg"/(24.305 cancel"g Mg") xx ("1 mol H"_2)/(cancel"1 mol Mg")`

`= "0.2467 mols H"_2` were produced.

Now, at this point, we would assume that `"H"_2(g)` is an ideal gas at `25^@ "C"` and `"1 atm"`, and use the ideal gas law:

`PV = nRT`

where:

• `P` is the pressure in `"atm"`.
• `V` is the volume in `"L"`.
• `n` is the `bb"mols"` of ideal gas.
• `R = "0.082057 L"cdot"atm/mol"cdot"K"` is the universal gas constant with the appropriate units.
• `T` is the temperature in `"K"`.

And thus,

`V_(H_2(g)) = (nRT)/P`

`= (("0.2467 mols H"_2)("0.082057 L"cdot"atm/mol"cdot"K")(25 + "273.15 K"))/("1 atm")`

`= "6.0396 L H"_2`

Unfortunately, you have only allowed yourself a measly one significant figure. :( I'm sorry.

You can only say you have `color(blue)("6 L H"_2)`...