Question

5 g of $CaCO_3$ completely reacts with

• A. 36.4 g of HCl
• B. 3.65 g of HCI
• C. 0.365 g of HCI
• D. 7.30 g of HCI.

Answer 1

Answer: (B)

3.65 g of HCI

Answer 2

The balanced equation for the reaction is:
$\underset{\text{100 g}}{{Ca}}$ $C0_3$ +$\underset{\text{73 g}}{{2H}}$ $CaCl_2 +CO_2+ H_2O$
100 g of $CaCO_3$ reacts completely with 73 g of HCl
So, 5 g of $CaCO_3$ will react with
$\frac{73}{100} \times 5 \, g$ of HCl
= 3.65 g of HCI