Chemistry problem: how to do, thanks?

a)
mass of 1 mole of MnO₂ = 54.9+2(16) = 86.9 g
mass of 1 mole of HCl = 1+35.5 = 36.5 g
217 g of MnO₂ contains: 217/86.9=2.497 moles
274 g of ‬HCl contains: 274/36.5= 7.507 moles

In this reaction, 
1 mole of MnO₂ reacts with 4 moles of HCl ,
y moles of MnO₂ will react with 7.507 moles of HCl
∴ y = 7.507/4 = 1.877 (moles) of MnO₂ 

Since there are 2.497 moles of MnO₂, 
MnO₂ will be in excess.
So, HCl is the limiting reactant.

b) 4 moles of HCl produce 1 mole of Cl₂  
=>7.507 moles HCl produce 7.507/4=1.877 mole Cl₂  

mass of 1 mole of Cl₂ = 2(35.5) = 71 g
∴mass of Cl₂ produced=71(1.877)=133.3 g(to 1 dec.pl.)‬

c) no. of moles of MnO₂ used = 1.877 
∴no. of moles of MnO₂ in excess=2.497-1.877=0.62

∴mass of excess reagent (MnO₂) left after reaction
= 86.9 (0.62) = 53.88 g (to 2 dec.pl.)