91Ó°ÊÓ

A sample of magnesium was burnt in air to give mixture of \(\mathrm{MgO}\) and \(\mathrm{Mg}_{3} \mathrm{~N}_{2}\). The ash was dissolved in 60 meq. of \(\mathrm{HCl}\) and the resulting solution back titrated with \(\mathrm{NaOH} .12 \mathrm{meq}\). of \(\mathrm{NaOH}\) were required to reach the end point. An excess of \(\mathrm{NaOH}\) was then added and the solution distilled. The ammonia released was then trapped in 10 meq. of second acid solution. Back titration of this acid solution required 6 meq. of the base. Calculate percentage magnesium burnt to the nitride. \(\quad\) [Roorkee 1998] [Hint : Let moles of \(\mathrm{Mg}\) used for \(\mathrm{MgO}\) and \(\mathrm{Mg}_{3} \mathrm{~N}_{2}\) are \(x\) and \(\mathrm{y}\) respectively. \(2 \mathrm{Mg}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{MgO}\) \(3 \mathrm{Mg}+\mathrm{N}_{2} \longrightarrow \mathrm{Mg}_{3} \mathrm{~N}_{2}\) Moles of \(\mathrm{MgO}=x\), Moles of \(\mathrm{Mg}_{3} \mathrm{~N}_{2}=y / 3\) \(\mathrm{MgO}+2 \mathrm{HCl} \longrightarrow \mathrm{MgCl}_{2}+\mathrm{H}_{2} \mathrm{O}\) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}+8 \mathrm{HCl} \longrightarrow 3 \mathrm{MgCl}_{2}+2 \mathrm{NH}_{4} \mathrm{Cl}\) tion will contain \(x\) moles of \(\mathrm{MgCl}_{2}\) and \(2 y / 3\) moles of \(\mathrm{NH}_{4} \mathrm{Cl}\). Moles of \(\mathrm{HCl}\) used \(=2 x+8 y / 3\) for \(\mathrm{MgO}\) and \(\mathrm{Mg}_{3} \mathrm{~N}_{2}\) \(2 x+\frac{8 y}{3}=60-12=48 \quad \ldots\) (i) Moles of \(\mathrm{NH}_{4} \mathrm{Cl}\) formed \(=\) moles of \(\mathrm{NH}_{3}\) \(=\) moles of HCl used to absorb \(\mathrm{NH}_{3}\) \(2 y / 3=4\) \(y=6\) From eq. (i) \(2 x+\frac{8 \times 6}{3}=48\) 5 \(\% \mathrm{Mg}\) used for \(\mathrm{Mg}_{3} \mathrm{~N}_{2}\) Page 1

Step by step solution

01

Understand the Reaction and Equations

When magnesium (Mg) burns in air, it forms both magnesium oxide (MgO) and magnesium nitride (Mg_3N_2). The burners yield x moles of MgO and y moles of Mg_3N_2. From the reaction equations, we know: \(2\mathrm{Mg} + \mathrm{O}_2 \rightarrow 2 \mathrm{MgO}\) for MgO and \(3\mathrm{Mg} + \mathrm{N}_2 \rightarrow \mathrm{Mg}_3\mathrm{N}_2\) for Mg_3N_2. The moles of MgO are x, and moles of Mg_3N_2 are y/3.

02

Chemical Reaction with HCl

The ash (consisting of MgO and Mg_3N_2) is treated with HCl. MgO reacts to form \(\mathrm{MgCl}_2\) consuming 2 moles of HCl per mole of MgO. Mg_3N_2 reacts to form both \(\mathrm{MgCl}_2\) and \(\mathrm{NH}_4\mathrm{Cl}\), consuming 8 moles of HCl per mole of Mg_3N_2. Therefore, total moles of HCl used are \(2x + \frac{8y}{3}\).

03

Back Titration with NaOH

The resulting solution is back titrated with NaOH, using 12 meq. After subtracting this from the initial 60 meq of HCl, \(48\) meq of HCl were utilized in the reaction: \(2x + \frac{8y}{3} = 48\).

04

Determine Moles of NH4Cl and NH3

The distilled ammonia, released from \(\mathrm{NH}_4\mathrm{Cl}\), is absorbed by another acid solution that takes up 4 meq. This relates to moles of \(\mathrm{NH}_4\mathrm{Cl}\), which equals moles of NH3. We have \(\frac{2y}{3} = 4\), solving for y gives \(y = 6\).

05

Solve for X using HCl Equation

Substitute y in the equation for HCl \(2x + \frac{8 \times 6}{3} = 48\). Simplifying, we find: \(2x + 16 = 48\), leading to \(2x = 32\), giving \(x = 16\).

06

Calculate Percentage of Mg for Mg3N2

Moles of Mg used in MgO are \(2x = 32\). Moles of Mg used in Mg_3N_2 are \(3y = 3\times 6 = 18\). Therefore, percentage of Mg used for Mg_3N_2 is \(\left(\frac{18}{32 + 18}\right) \times 100\% = 36\%\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnesium Oxide (MgO)

Magnesium oxide, or MgO, is an essential compound that forms when magnesium burns in the presence of oxygen. This process involves the combination of magnesium atoms with oxygen molecules, resulting in the production of MgO. The chemical reaction can be written as:

• \(2\, \text{Mg} + \text{O}_2 \rightarrow 2\, \text{MgO}\)

This reaction is significant because MgO is commonly found in nature as the mineral periclase. It is highly stable and exhibits basic properties, which means it can react with acids such as hydrochloric acid (HCl) to form magnesium chloride (MgClâ‚‚) and water. It's crucial to understand that this transformation consumes two moles of HCl for every mole of MgO present, thus:

• \(\text{MgO} + 2\,\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\text{O}\)

MgO's ability to easily form MgClâ‚‚ upon reacting with HCl is important in titration techniques, where the complete neutralization of an acidic solution is required.

Magnesium Nitride (Mg3N2)

Mg₃N₂, or magnesium nitride, is another product formed when magnesium burns in an atmosphere with nitrogen. This compound is far less common than MgO due to the lower reactivity of nitrogen compared to oxygen. The chemical equation for this reaction is:

• \(3\,\text{Mg} + \text{N}_2 \rightarrow \text{Mg}_3\text{N}_2\)

Upon formation, magnesium nitride showcases unique chemical properties. When it reacts with hydrochloric acid, it forms magnesium chloride and ammonium chloride (NH₄Cl). This reaction highlights Mg₃N₂'s capability to produce ammonia, which can be distilled and collected:

• \(\text{Mg}_3\text{N}_2 + 8\,\text{HCl} \rightarrow 3\,\text{MgCl}_2 + 2\,\text{NH}_4\text{Cl}\)

The generated ammonia can then be neutralized using another acid, a crucial step in calculating the original composition of magnesium nitride in a sample. This reactions system is a simultaneous demonstration of chemical synthesis and analysis, providing insightful understanding into elemental reactions and compound properties.

Back Titration

Back titration is a valuable method in analytical chemistry that involves titrating the remaining amount of a reactant following an initial reaction. It is particularly useful when direct titration is impractical.
In the context of the exercise, after burning magnesium and forming both MgO and Mg₃N₂, hydrochloric acid is used to fully dissolve the resulting ash. Initially, the excess hydrochloric acid left after the reaction with MgO and Mg₃N₂ is titrated with sodium hydroxide (NaOH). Back titration helps calculate the amount of acid that did not react, as only 48 meq out of 60 meq is consumed:

• This is achieved through: \(60 - 12 = 48\)
• The equation for HCl usage in MgO and Mg₃Nâ‚‚: \(2x + \frac{8y}{3} = 48\)

Following this, any ammonia released via the reaction with Mg₃N₂ is recovered and titrated using more acid.
This means that by knowing the amount of unreacted acid after ammonia absorption, these details can reconstruct the reaction profile accurately from the remaining acid quantities. The recovery and titration of ammonia offer a form of cross-verification for the quantities of magnesium nitride initially present.