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

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)#...