Question

X, Y and Z are three crystalline solids that are soluble in water and have a common anion. To help you to identify X, Y and Z, you are provided with the following experimental observations. A reddish-brown gas is obtained when X, Y and Z are separately warmed with concentrated sulphuric acid and copper turnings added to the mixture.

Answer 1

Analysis of given conditions:

• X, Y, and Z are water-soluble crystalline solids with a common anion.
• A reddish-brown gas's evolution occurs when they are warmed separately with concentrated Sulphuric acid(${\mathrm{H}}_2{\mathrm{SO}}_4$) and Copper turnings are added to the mixture.
• It can be inferred that the common anion is Nitrate (${{\mathrm{NO}}_3}^{-}$).

Chemical reactions involved:

• When Nitrate is heated with concentrated Sulphuric acid below 200-degree Celsius, a reddish-brown colored Nitric acid (${\mathrm{HNO}}_3$) is evolved.

${{\mathrm{NO}}_3}^{-}\left(\mathrm{l}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\stackrel{<200°\mathrm{C}}{\to }{{\mathrm{HSO}}_4}^{-}\left(\mathrm{l}\right)+{\mathrm{HNO}}_3\left(\mathrm{g}\right)$

(Nitrate ion) (Bisulphate ion) (Reddish-brown gas)

• When X is heated, a gas is evolved participating in the glowing of a splint. Hence. X is Potassium nitrate(${\mathrm{KNO}}_3$).
• Potassium nitrate (${\mathrm{KNO}}_3$) decomposes into Potassium nitrite(${\mathrm{KNO}}_2$) and Oxygen(${\mathrm{O}}_2$) upon heating.

$2{\mathrm{KNO}}_3\left(\mathrm{s}\right)\stackrel{∆}{\to }2{\mathrm{KNO}}_2\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$

(Potassium (Potassium

nitrate) nitrite)

• When Y is heated, a reddish-brown gas is evolved and a yellow residue fuses with the test tube glass. Hence, Y is Lead nitrate($\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$).
• Lead nitrate ($\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$) decomposes to yellow residue i.e. Lead oxide($\mathrm{PbO}$), reddish-brown gas i.e. Nitrogen dioxide(${\mathrm{NO}}_2$) and Oxygen(${\mathrm{O}}_2$) upon heating.

$2\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)\stackrel{∆}{\to }2\mathrm{PbO}\left(\mathrm{s}\right)+4{\mathrm{NO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$

(Lead nitrate) (Lead oxide) (Nitrogen dioxide)

• When Z is warmed with Sodium hydroxide solution($\mathrm{NaOH}$), a gas is liberated, turning moist red litmus blue. Hence, Z is Ammonium nitrate(${\mathrm{NH}}_4{\mathrm{NO}}_3$).
• When Ammonium nitrate(${\mathrm{NH}}_4{\mathrm{NO}}_3$) is warmed with Sodium hydroxide solution($\mathrm{NaOH}$), Sodium nitrate(${\mathrm{NaNO}}_3$) is formed along with the liberation of Ammonia (${\mathrm{NH}}_3$) which is basic in nature.

${\mathrm{NH}}_4{\mathrm{NO}}_3\left(\mathrm{s}\right)+\mathrm{NaOH}\left(\mathrm{aq}\right)\stackrel{∆}{\to }{\mathrm{NaNO}}_3\left(\mathrm{aq}\right)+{\mathrm{NH}}_3\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

(Ammonium (Sodium (Sodium (Ammonia)

nitrate) hydroxide) nitrate)

Hence, X is Potassium nitrate(${\mathrm{KNO}}_3$), Y is Lead nitrate($\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$), and Z is Ammonium nitrate(${\mathrm{NH}}_4{\mathrm{NO}}_3$