Highly selective Backlog Qs for JEE MainChemistrySome Basic Concepts of Chemistry
100 ~mL of N 5 HCl was added to 1 ~g of pure CaCO ₃ . What would remain after the reaction?
Options
- A0.5 ~g of CaCO ₃
- BNeither CaCO ₃ nor HCl
- C50 ~mL of HCl
- D25 ~mL of HCl
Correct answer
B. Neither CaCO ₃ nor HCl
Step-by-step solution
Number of moles of CaCO ₃= W M = 1 100 =0.01 ~mol Molar mass of CaCO ₃=100 ~g ~mol ⁻¹100 ~mL of N 5 HCl = 100 1000 1 5 =0.02 ~mol The reaction is 0.01 mole of CaCO ₃ reacts with 0.02 mole of HCl . Both are completely consumed in the given reaction.