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##2.24 “L”##
Step 1: determine the reaction
The chemical equation is already given.
##”Mg” + 2 “HCl” -> “MgCl”_2 + “H”_2##
Step 2: determine which reactant is limiting
0.100 mol of Mg reacts with 0.500 mol of HCl, which of them will run out first?
From the reaction, the ratio of ##”Mg”## to ##”HCl”## used is ##1:2##. This means that if the reaction goes to completion, the ##0.100 “mol”## of ##”Mg”## will react with ##0.100 “mol” xx 2 = 0.200 “mol”## of ##”HCl”##. This leaves ##0.500 “mol” – 0.200 “mol” = 0.300 “mol”## of ##”HCl”##.
Step 3: determine how much ##”H”_2## is produced.
##0.100 “mol”## of ##”Mg”## and ##0.200 “mol”## of ##”HCl”## will produce ##0.100 “mol”## of ##”H”_2##.
Step 4: determine the volume at STP
Each mole of ideal gas occupies ##22.4 “L”## at STP. Therefore, ##0.100 “mol”## of ##”H”_2## will occupy ##2.24 “L”## of space.
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