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 )+2 HCl_ ( aq ) CaCl_2 ( aq )+ CO_2 ( g )+2H_2O_ (l ) [Calcula…

Question

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 )+2 HCl_ ( aq ) CaCl_2 ( aq )+ CO_2 ( g )+2H_2O_ (l ) [Calculate upto second place of decimal point]

Accepted Answer

Correct answer: C. (c) CaCO_3 ( s )+2 HCl_ ( aq . ) CaCl_2 ( aq . )+ CO_2 ( g )+H_2O_ (l ) no. of moles of CaCO_3 (pure) = 12 mole [No. of Moles = molarity volume of solution (in ltr.) ] = 12 0.5 501000=0.0125 weight of CaCO_3 (pure) = mole mol. wt =0.0125 100=1.25 g \% purity= wt . of pure substance wt . of impure sample 100 wt. of impure sample = 1.25 10095=1.32 g

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

Solution

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