Question

Mass of caco3 produced when carbon dioxide is passed in excess to 500 ml of 0.5 m ca[oh]2

Answer 1

Let's see the chemical reaction between Calcium hydroxide and carbon dioxide....

$Ca(OH)_2+CO_2\rightarrow CaCO_3+H_2O$

given, volume of Ca(OH)2 = 500mL = 0.5litre

concentration of Ca(OH)2 = 0.5M

so, number of mole of Ca(OH)2 = volume in Litre × concentration

= 0.5 Litre × 0.5M

= 0.25 mole

from reaction,

1 mole of Ca(OH)2 produces 1 mole of CaCO3.

so, 0.25 mole of Ca(OH)2 produces 0.25 mole of CaCO3.

hence, mole of CaCO3 = 0.25 mol

mass of CaCO3 = mole of CaCO3 × molar mass of CaCO3

= 0.25mol × 100g/mol

= 25g

hence, mass of CaCO3 = 25g

Answer 2

Let's see the chemical reaction between Calcium hydroxide and carbon dioxide....

Ca(OH)_2+CO_2

CaCO_3+H_2OCa(OH)2+CO2→CaCO3+H2O

given, volume of Ca(OH)2 = 500mL = 0.5litre

concentration of Ca(OH)2 = 0.5M

so, number of mole of Ca(OH)2 = volume in Litre × concentration

= 0.5 Litre × 0.5M

= 0.25 mole

from reaction,

1 mole of Ca(OH)2 produces 1 mole of CaCO3.

so, 0.25 mole of Ca(OH)2 produces 0.25 mole of CaCO3.

hence, mole of CaCO3 = 0.25 mol

mass of CaCO3 = mole of CaCO3 × molar mass of CaCO3

= 0.25mol × 100g/mol

= 25g

hence, mass of CaCO3 = 25g