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 + 2 H 2 O l [Calculate upto second place of decimal point]

Step-by-step solution

( (s) CaCO ₃ + 50 ~mL , 0.5 M 2 HCl ( aq ) ( aq ) CaCl ₂ + (g) CO ₂ + (l) 2 H ₂ O ) the number of moles of HCl taken (=0.5 0.05 ) (=0.025 ) moles As we can see that from the above balanced equation is one mole of ( CaCO ₃(s) ) requires 2 moles of ( HCl ( aq ) ) therefore, for 0.025 moles of ( HCl ( aq ), 0.0125 ) moles of ( CaCO ₃(s) ) will be required. Mass of 0.0125 moles of ( CaCO ₃(s)=0.0125 ) molar mass of ( CaCO ₃ ) ( aligned & =0.0125 100 & =1.25 ~g aligned ) But purity of ( CaCO ₃(s) ) is (95 % ). Therefore