This reaction will produce #"2.63 g"# of #Cl_2# gas.
Start with the balanced chemical equation, that is a must for any stoichiometry problem.
#MnO_(2(s)) + 4HCl_((aq)) -> Cl_(2(g)) + MnCl_(2(aq)) + 2H_2O_((l))#
Now look at the mole ratios you have between #MnO_2#, #HCl#, and #Cl_2#: 1 mole of #MnO_2# needs 4 moles of #HCl# in order to produce 1 mole of #Cl_2# gas.
SInce you can calculate how many moles of both #MnO_2# and #HCl# you have, you need to determine whether or not one of them acts as a limiting reagent, i.e. is in insufficient amount when compared with the other one.
#"5.00 g" * ("1 mole")/("87.0 g") = "0.0575 moles"# #MnO_2#
Use the molarity of the #HCl# solution to determine how many moles you have
#C = n/V => n_(HCl) = C * V = "6.0 M" * 25 * 10^(-3)"L" = "0.15 moles HCl"#
However, according to the mole ratio, you would have needed
#"0.0575 moles MnO"_2 * ("4 moles HCl")/("1 mole MnO"_2) = "0.0230 moles HCl"#
This means that #HCl# is acting as a limiting reagent and will determine how much #MnO_2# actually reacts
#"0.15 moles HCl" * ("1 mole MnO"_2)/("4 moles HCl") = "0.0375 moles"# #MnO_2#
This will be equal to the number of #Cl_2# moles produced - remember the #"1:1"# mole ratio between the two compounds. Therefore, the mass of #Cl_2# gas will be
#"0.0375 moles" * ("70.0 g")/("1 mole") = "2.63 g"#
You can adjust this to the number of sig figs you actually have, since I think 6 M was actually 6.0, or 6.00 M.
SIDE NOTE The net ionic equation for this reaction is
#MnO_(2(s)) + 4H_((aq))^(+) + 2Cl_((aq))^(-) -> Cl_(2(g)) + Mn_((aq))^(2+) + 2H_2O_((l))#
