(50 ~g CaCO ₃ ) is allowed to react with (49 ~g H ₃ PO ₄ ) according to the given reaction - (3 CaCO ₃+2 H ₃ PO ₄ Ca ₃ ( PO ₄ )₂+3 H ₂ O +3 CO ₂ ) Then pick out the correct option (A) 51.67 g of salt is formed (B) Amount of unreacted reagent (=32.67 ~g ) (C) Moles of ( CO ₂ ) formed (=0.5 ) moles (D) Mass of ( H ₂ O ) formed (=27 ~g )

Step-by-step solution

Given Reaction: (3 CaCO ₃+2 H ₃ PO ₄ Ca ₃ ( PO ₄ )₂+ ) Molar Masses - ( CaCO ₃: 100 ~g / mol ) - ( H ₃ PO ₄: 98 ~g / mol ) Moles of Reactants - Moles of ( CaCO ₃ ) : ( aligned & 50 ~g 100 ~g / mol =0.5 moles & 49 ~g 98 ~g / mol =0.5 moles aligned ) Limiting Reagent From the stoichiometry: - Required: 3 moles of ( CaCO ₃ ) for 2 moles of ( H ₃ PO ₄ ). - Since both reactants are present in a 1:1 ratio, neither is limiting for the given amounts. Products Formed - Moles of ( CaCO ₃ ) used (=0.5 ) moles ( ) Produces 0.1