Please answer as soon as possible...
Question:9
MnO₂ + 4HCl $\to$ MnCl₂+ 2H₂O + Cl₂
1.746g of pure MnO₂ is heated strongly with conc. HCl. Calculate:
a) Moles of MnO₂ used.
b) Mass of salt formed.
c) Moles of chlorine gas formed.
d) Volume of chlorine gas formed at STP.
e) Mass of acid required.

Dear student ,

a) Moles of MnO₂ used = Given mass/ molecular mass
1.746/86.94 = 0.02 moles

b) According to equation 1 mole of MnO₂ produces 1 mole of salt MnCl₂ 
0.02 mole of MnO₂ produces 0.02 mole of MnCl₂

c)  According to equation 1 mole of MnO₂ produces 1 mole of chlorine gas
0.02 mole of MnO₂ produces 0.02 mole of chlorine gas formed

d) 1 mole of chlorine gas occupies 22.4 L at STP
0.02 mole of chlorine gas occupies 22.4$\times$0.02 = 0.448 L of Cl₂ gas 

e) Moles of acid required = 0.08 moles
Mass of HCl = mole $\times$ Molecular mass
0.08 $\times$ 36.5 = 2.92 g 

Regards