What mass of 95% pure $CaCO_3$ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? $CaCO_{3(s)} + 2HCl_{(aq)} \rightarrow CaCl_{2(aq)} + CO_{2(g)} + 2H_2O_{(l)}$ [Calculate upto second place of decimal point]

Question

Dr. Abubacker · Accepted Answer

[Solution] Moles of HCl = $0.05 	imes 0.5 = 0.025$; moles of $CaCO_3 = 0.0125$, mass $= 1.25$ g of pure $CaCO_3$. Impure sample needed $= 1.25/0.95 = 1.32$ g. Answer (3).

Q. What mass of 95% pure $CaCO_3$ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? $CaCO_{3(s)} + 2HCl_{(aq)} \rightarrow CaCl_{2(aq)} + CO_{2(g)} + 2H_2O_{(l)}$ [Calculate upto second place of decimal point]

Explanation

Reviewed by Dr. Abubacker

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