#MnO_2 + 4HCl -> MnCl_2 + 2H_2O + Cl_2#

We are only considering #MnO_2 + 4HCl# since the question deals only with the reactants and not the products..

Stoichiometric ratio #-># #1:4#

Since #1 "mole"# of #MnO_2# reacts with #4 "moles"# of #HCl#

Therefore #x# moles of #MnO_2# will react with 2.5 moles of #HCl#

#rArr# #1/x = 4/2.5#

#rArr# #4 xx x = 2.5 xx 1#

#rArr# #4x = 2.5#

#rArr# #x = 2.5/4#

#rArr# #x = 0.625#

Hence no of moles of #MnO_2 = 0.625mols#

**Recall** #-># #"No of moles" = "Mass"/"Molar mass"#

#rArr# #"Mass" = "No of moles" xx "Molar Mass"#

#"Molar Mass of" color(white)(x) MnO_2 = 25 + (16 xx 2)#

#"Molar Mass of" color(white)(x) MnO_2 = 25 + 32#

#"Molar Mass of" color(white)(x) MnO_2 = 57gmol^-1#

#"Mass of" color(white)(x) MnO_2= "No of moles of" color(white)(x) MnO_2 xx "Molar Mass of" color(white)(x) MnO_2#

#"Mass of" color(white)(x) MnO_2= 0.625mols xx 57gmol^-1#

#"Mass of" color(white)(x) MnO_2= 0.625cancel(mols) xx 57gcancel(mol^-1)#

#"Mass of" color(white)(x) MnO_2= 35.625g#

#"Mass of" color(white)(x) MnO_2~~ 36g#