a. Starting with calcium chloride, describe briefly how a solid sample of calcium trioxocarbonate (IV)...
a. Starting with calcium chloride, describe briefly how a solid sample of calcium trioxocarbonate (IV) can be prepared in the laboratory.
b. With relevant equations outline the procedure for the purification of impure copper.
c. Copper reacts with concentrated trioxonitrate (V) acid: i. write a balanced chemical equation for the reaction;
ii. state what would be observed in the reaction,
iii. state why the copper is oxidized;
iv. an excess of copper is added to 25.0 \(cm^3\) of 16.0 mol \(dm^3\) \(HNO_3\). Calculate the volume of the gas formed at s.t.p. [H=1.0, N=14.0, O= 16.0, Cu=63.0; Molar volume of gas at .s.t.p.=22.4 \(dm^3\)]
di. Pure \(HNO_3\), is a colourless liquid but when exposed to air, it turns yellowish-brown in colour. Explain briefly this observation.
ii. Write a balanced equation for the laboratory preparation of hydrogen trioxonitrate (V) acid.
Explanation
a. Water is added to the \(CaCl_2\): to form a solution. \(Na_2CO\), is then added to the solution to precipitate \(CaCO_3\),which is filtered, washed, and dried.
b. Electricity is passed through a solution of CuSO4 using an impure copper as the anode and pure copper as the cathode. During electrolysis,the anode loses mass as copper dissolves and the cathode gains mass as copper is deposited.
Anode: \(Cu_s → Cu^{2+}_{(aq)} + 2e^-\)
Cathode: \(Cu^{2+}_{(aq)} + 2e^- → Cu_{(s)}\)
ci. \(Cu_{(s)} + 4HNO_{3(aq)} →Cu(NO_3)_{2(aq)} + 2NO_{2(g)} + 2H_2O_{(l)}\)
ii. -blue solution
-(reddish)brown gas
-gas bubbles/effervescence
-metal dissolves/metal disappears
iii. Because copper loses electrons/oxidation number increases.
iv. Moles of acid=0.025x16= 0.4 mole
Mole of \(NO_2\) = moles of acid/2 =0.2 mole
Volume of gas (\(NO_2)\) at s.t.p. = 0.2 x 22.4 \(dm^{-3}\) = 4.48 mol \(dm^{-3}\)
di. Pure HNO3 undergoes decomposition to \(NO_2\)(brown gas). This gas dissolves in the rest of the acid leaving it with that colour.
ii. -2\(KNO_3 + H_2SO_4→K_2SO_4+2HNO_3\)
- \(NaNO_3 + H_2SO_4→NaHSO_4 + HNO_3\)
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