A sample of hard water which contain equimolar mixture of Ca ( HCO 3 ) 2 and Mg ( HCO 3 ) 2 , is treated with 3000 mL , 0.5 M Ca ( OH ) 2 solution such that none of the reactants are left. A

Question

A sample of hard water which contain equimolar mixture of Ca ( HCO 3 ​ ) 2 ​ and Mg ( HCO 3 ​ ) 2 ​ , is treated with 3000 mL , 0.5 M Ca ( OH ) 2 ​ solution such that none of the reactants are left. Amount of precipitate formed (in grams) will be ____ . [Fill your answer as sum of digits (excluding decimals places) till you get single digit answer] (Consider all Mg 2+ to be precipitated as Mg(OH) 2 and all Ca 2+ as CaCO 3 ) (Ca = 40, C = 12, Mg = 24, O = 16, H = 1)

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Moles of Ca(OH)₂ used: M o l a r i t y × V o l u m e = 0.5 M × 3 L = 1.5moles . Reaction Stoichiometry: Let x be the moles of each in the equimolar mixture: C a ( H C O 3 ​ ) 2 ​ + C a ( O H ) 2 ​ → 2 C a C O 3 ​ ↓ + 2 H 2 ​ O ( re q u i res x m o l eso f C a ( O H ) 2 ​ ) M g ( H CO _ 3 ) _ 2 + 2 C a ( O H ) _ 2 → M g ( O H ) _ 2 ↓ + 2 C a CO _ 3 ↓ + 2 H _ 2 O ( re q u i res 2 x m o l eso f C a ( O H ) _ 2 ) Find x : Total Ca(OH)₂ = x + 2 x = 3 x = 1.5 ⟹ x = 0.5mole . Mass of Precipitates: T o t a lC a C O 3 ​ = 4 x = 2 m o l es → 2 × 100 g / m o l = 200 g T o t a lM g ( O H ) 2 ​ = x = 0.5 m o l e → 0.5 × 58 g / m o l = 29 g Total weight & Digit Sum: T o t a lM a ss = 200 + 29 = 229 g . Summing the digits: 2 + 2 + 9 = 13 → 1 + 3 = 4 .

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