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

##### 1 Answer

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

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

If we are on the same page, then **for every** *consumed*, *made*. This was found from the stoichiometry of the reaction, i.e.

This means that from the molar mass of

#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# wereproduced.

Now, at this point, we would assume that **ideal gas law**:

#PV = nRT# where:

#P# is thepressurein#"atm"# .#V# is thevolumein#"L"# .#n# is the#bb"mols"# of ideal gas.#R = "0.082057 L"cdot"atm/mol"cdot"K"# is theuniversal gas constantwith the appropriate units.#T# is thetemperaturein#"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