(a) State Gay Lussac's Law. (b)… | Chemistry

Question 1 Report

(a) State Gay Lussac's Law.

(b) Carbon (II) oxide reacted with oxygen to form carbon (IV) oxide in a see tube.

(i) Write a balanced equation for the reaction.

(ii) If 40 cm\(^3\) of the carbon (II) oxide were mixed with cm\(^3\) of oxygen,

I. calculate the volume of carbon (IV) oxide produced

II. which reactant is in excess and how much?

III. what was the total volume of the gaseous mixture at the end of the reaction?

(i) is an acidic oxide that is insoluble in water?

(ii) reacts with water to give alkaline solution?

(iii) is amphoteric?

(iv) is neutral?

(v) is/are gaseous at room temperature?

(d) Explain why

(i) colourless concentrated trioxonitrate (V) acid turns yellow,

(ii) dilute trioxonitrate (V) acid does not liberate hydrogen when it reacts with magnesium.

(e) Write a chemical equation for the thermal decomposition of (i) Cu(NO\(_3\))\(_{2(g)}\)

(ii) NH\(_4\)NO\(_{3(g)}\)

Answer

(a) Gay-Lussac's law (of combining volumes)

When gases react, they do so in volumes which bear a simple whole-number ratio to one another and to the volumes of the gaseous products, provided the temperature and pressure remain constant.

(b)(i) Balanced equation

\[2CO_{(g)} + O_{2(g)} \to 2CO_{2(g)}\]

(b)(ii) Volume calculation

The stem gives 40 cm\(^3\) of carbon(II) oxide but the volume of oxygen is missing from the text; the working below is shown for the common case of 40 cm\(^3\) of oxygen and must be redone with the correct oxygen volume.

From \(2CO : 1O_2 : 2CO_2\), 40 cm\(^3\) CO reacts with 20 cm\(^3\) \(O_2\) to give 40 cm\(^3\) \(CO_2\).

I. Volume of \(CO_2\) produced: 40 cm\(^3\).
II. Reactant in excess: oxygen; \(40 - 20 = 20\,cm^3\) of oxygen is left over.
III. Total volume at the end: \(40\,(CO_2) + 20\,(\text{excess } O_2) = 60\,cm^3\).

(c) The oxides CaO, SiO2, CO, NO2, ZnO

(i) Acidic oxide insoluble in water: \(SiO_2\).
(ii) Reacts with water to give an alkaline solution: \(CaO\).
(iii) Amphoteric: \(ZnO\).
(iv) Neutral: \(CO\).
(v) Gaseous at room temperature: \(CO\) and \(NO_2\).

(d) Explanations

(i) Colourless concentrated trioxonitrate(V) acid turns yellow because it partly decomposes, especially in light and warmth, releasing brown nitrogen(IV) oxide which dissolves in the acid: \(4HNO_3 \to 4NO_2 + O_2 + 2H_2O\).
(ii) Dilute trioxonitrate(V) acid does not liberate hydrogen with magnesium because it is a strong oxidizing acid; it oxidizes any hydrogen produced to water while it is itself reduced to oxides of nitrogen, so hydrogen gas is not evolved.

(e) Thermal decomposition

\[\text{(i) } 2Cu(NO_3)_2 \to 2CuO + 4NO_2 + O_2\]\[\text{(ii) } NH_4NO_3 \to N_2O + 2H_2O\]

Read lesson note on Metals And Their Compounds (JAMB)

Read lesson note on Stoichiometry And Chemical Reactions (WAEC)

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