Question

Write a balanced equation for the dissolution of sodium carbonate (Na2CO3) in water. Find the number of moles of Na+ produced when 0.207 mol of sodium carbonate dissolves.

Answer 1

The balanced equation for the dissolution of sodium carbonate in water produces two moles of Na+ for every mole of Na2CO3. For 0.207 mol of sodium carbonate, it yields 0.414 mol of Na+ ions.

Step-by-step explanation:

The balanced equation for the dissolution of sodium carbonate (Na2CO3) in water is:

Na2CO3(s) → 2Na+(aq) + CO3^2-(aq)

When 0.207 mol of sodium carbonate dissolves in water, it produces two moles of Na+ ions for every mole of sodium carbonate. Therefore, to find the number of moles of Na+ produced we multiply:

0.207 mol Na2CO3 × 2 mol Na+ / 1 mol Na2CO3 = 0.414 mol Na+

Answer 2

in the presence of water H2O

Na2CO3 (S) --> 2Na+ (aq)+ (CO3)2-(aq)

One mole of sodium carbonate produces two moles of Na+ ions

Therefore 0.207 moles produces 0.414 moles of Na+ ions

= 0.414 moles of Na+ ions

Step-by-step explanation:

In water

Na2CO3 --> 2Na+ (aq)+ (CO3)2-(aq)

In a limited reaction, the carbonate ion reacts with the water molecules as follows

(CO3)2-(aq) + H2O←→HCO3-(aq) + OH-(aq)

sodium carbonate or soda ash dissolves in water to give 2 sodium cations and one carbonate anion