Question

Write the balanced chemical equations for the following reactions:

(a) Sodium carbonate on reaction with hydrochloric acid in equal molar concentrations gives sodium chloride and sodium hydrogen carbonate.

(b) Sodium hydrogen carbonate on reaction with hydrochloric acid gives sodium chloride, water liberates carbon dioxide.

(c) Copper sulfate on treatment with potassium iodide precipitates cuprous iodide, liberates iodine gas, and also forms potassium sulfate.

Answer 1

Step 1:

Chemical equations:

Chemical equations are the symbolic representation of chemical reactions. It involves chemical formulae and chemical symbols of the compounds.

Step 2:

The balanced chemical equations are given as:

(a)

${\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right)+\mathrm{HCl}\left(\mathrm{aq}\right)\to \mathrm{NaCl}\left(\mathrm{aq}\right)+{\mathrm{NaHCO}}_3\left(\mathrm{aq}\right)$

(Sodium (Hydrochloric (Sodium (Sodium

carbonate) acid) chloride) bicarbonate)

From the balanced equation, 1 mole of Sodium carbonate (Na₂CO₃) reacts with 1 moles of Hydrochloric acid (HCl) solution to form 1 mole each of Sodium chloride (NaCl) and Sodium bicarbonate (NaHCO₃).

(b)

${\mathrm{NaHCO}}_3\left(\mathrm{aq}\right)+\mathrm{HCl}\left(\mathrm{aq}\right)\to \mathrm{NaCl}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

(Sodium (Hydrochloric (Sodium (Water) (Carbon

bicarbonate) acid) chloride) dioxide)

From the balanced equation, 1 mole of Sodium bicarbonate (NaHCO₃) reacts with 1 mole of Hydrochloric acid (HCl) to form 1 mole each of Sodium chloride, Water (H₂O) and Carbon dioxide (CO₂),

(c)

$2{\mathrm{CuSO}}_4\left(\mathrm{aq}\right)+4\mathrm{KI}\left(\mathrm{aq}\right)\to {\mathrm{Cu}}_2{\mathrm{I}}_2\left(\mathrm{s}\right)+2{\mathrm{K}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+{\mathrm{I}}_2\left(\mathrm{g}\right)$

(Copper (Potassium (Cuprous (Potassium (Iodine)

sulfate) iodide) iodide) sulfate)

From the balanced equation, 2 moles of Copper sulfate (CuSO₄) reacts with 4 moles of Potassium iodide (KI) to form 1 mole of Cuprous iodide (Cu₂I₂), 2 moles of Potassium sulfate (K₂SO₄) and 1 mole of Iodine (I₂).