Question

If 20 g of $CaCO_3$ is treated with 100 mL of 20% HCl solution, the amount of $CO_2$ produced is

• A. 22.4 L
• B. 8.80 g
• C. 4.40g
• D. 2.24 L

Answer 1

Answer: (B)

8.80 g

Answer 2

$\underset{\text{100 g}}{ {CaCO_{3}}}+ \underset{\text{73 g}}{ {2 HCl}} \to CaCl+ \underset{\text{44 g}}{ {CO_{2}}}+H_{2}O$ $100\, mL$ of $20\%$ $HCl$ solution $= 20\, g \,HCl$ Here $CaCO_{3}$ is the limiting reactant $100\,g$ of $CaCO_{3}$ gives $44\,g\,CO_{2}$ $20\,g \, CaCO_{3}$ gives $\frac{44}{100}\times20$ $=8.80\,g$ of $CO_{2}$