100 mL of N / 5 HCl was added to 1 g of pure CaCO 3 What would remain after the reaction?

Question

100 mL of N / 5 HCl was added to 1 g of pure CaCO 3 What would remain after the reaction? (A) 0.5 g of of CaCO3 (B) Neither CaCO3 nor HCI (C) 50 mL of HCI (D) 25 mL of HCI

Tardigrade Admin · Accepted Answer

No. of moles of CaCO 3=(w/M)=(1/100) =0.01 mol [∵. Molar mass of .CaCO 3=100 g mol -1] 100 mL of N / 5 HCl =(100/1000) × (1/5) =0.02 mol The reaction is represented as: CaCO 3+2 HCl longrightarrow CaCl 2+ CO 2+ H 2 O 1 mole of CaCO 3 reacts with 2 moles of HCl. Hence, 0.01 mole of CaCO 3 will react completely with 0.02 mole of HCl.

Q. $100 m L$ of $N /5 H Cl$ was added to $1 g$ of pure $C a C O_{3}$ What would remain after the reaction?

0.5 g of of $C a C O_{3}$

Neither $C a C O_{3}$ nor HCI

50 mL of HCI

25 mL of HCI

Q. 100mL of N/5HCl was added to 1g of pure CaCO3​ What would remain after the reaction?

0.5 g of of CaCO3​

Neither CaCO3​ nor HCI

50 mL of HCI

25 mL of HCI

Q. $100 m L$ of $N /5 H Cl$ was added to $1 g$ of pure $C a C O_{3}$ What would remain after the reaction?

0.5 g of of $C a C O_{3}$

Neither $C a C O_{3}$ nor HCI