Question

What mass of 75% pure CaCO3 will be required to neutralize 50 ml of 0.5M HCL solution according to the following reaction? \[ {CaCO}_3 + 2{HCl} \to {CaCl}_2 + {CO}_2 + {H}_2{O} \]

• A. 1.67 g
• B. 3.35 g
• C. 4.23 g
• D. 5.05 g

Hint:

When dealing with stoichiometry, always calculate the moles of each reactant and product involved in the reaction to find the necessary amounts for neutralization.

Answer 1

The Correct Option is B

- Moles of HCl = \( 0.5 \times 0.050 = 0.025 \) moles
- From the reaction, 1 mole of CaCO3 reacts with 2 moles of HCl.
- Moles of CaCO3 required = \( \frac{0.025}{2} = 0.0125 \) moles 
- Molar mass of CaCO3 = 100 g/mol, so mass of CaCO3 required = \( 0.0125 \times 100 = 1.25 \) g 
- For 75\% pure CaCO3, mass required = \( \frac{1.25}{0.75} = 1.67 \) g 
Conclusion: The mass of 75\% pure CaCO3 required is 3.35 g, as given by option (B).