Q.What is the mass of $95\%$ pure $CaCO_3$ required to neutralize $50 \text{ mL}$ of $0.5 \text{ M } HCl$ solution according to the reaction: $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)$ ?

1.25 g

1.32 g

3.65 g

9.50 g

The correct solution involves applying the fundamental concept to derive the final value step by step...

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