Reaction: Ca+2HCl→CaCl2+H2
Moles of Ca = 4014.0=0.35 mol. Since HCl is in excess, Ca is limiting.
Option (1): H2 evolved = 0.35 mol
Option (2): Volume of H2 at STP = 0.35×22.4=7.84 L
Option (3): Ca is limiting reagent
Option (4): Mass of CaCl2=0.35×111=38.85 g, not 33.3 g
Statement (4) is incorrect.