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))