Question

What will be the amount of lime required to remove the hardness in 60l ofpond water containing 1.62mg of calcium bicarbonate per 100ml of water?

Answer 1

1.62 mg of Calcium bicarbonate Ca(HCO₃)₂ per 100 ml of water .
∴ amount of CA(HCO₃)₂ in 60L water = 1.62 × 10⁻³ g/100ml × 60× 1000mL
= 1.62 × 6/10 = 0.972g

Now , mole of Ca(HCO₃)₂ = weight/molar weight
= 0.972/162 [ because molar mass of calcium bicarbonate = 162g/mol ]
= 0.006 mole

Reaction is ------> Ca(HCO₃)₂ ⇔ Ca(OH)₂ + 2CO₂
∵ 1 mole of bicarbonate forms 1 mole of lime
∴ 0.006 mole of bicarbonate forms 0.006 mole of lime

∴ mole of lime = 0.006
0.006 = weight of lime/molar weight of lime
0.006 = weight of lime/74
weight of lime = 0.006 × 74 = 0.444g

Hence , weight of lime required to remove = 0.444g

Answer 2

Answer:

1.62 mg of Calcium bicarbonate Ca(HCO₃)₂ per 100 ml of water .

∴ amount of CA(HCO₃)₂ in 60L water = 1.62 × 10⁻³ g/100ml × 60× 1000mL

= 1.62 × 6/10 = 0.972g

Now , mole of Ca(HCO₃)₂ = weight/molar weight

= 0.972/162 [ because molar mass of calcium bicarbonate = 162g/mol ]

= 0.006 mole

Reaction is ------> Ca(HCO₃)₂ ⇔ Ca(OH)₂ + 2CO₂

∵ 1 mole of bicarbonate forms 1 mole of lime

∴ 0.006 mole of bicarbonate forms 0.006 mole of lime

∴ mole of lime = 0.006

0.006 = weight of lime/molar weight of lime

0.006 = weight of lime/74

weight of lime = 0.006 × 74 = 0.444g

Hence , weight of lime required to remove = 0.444g