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One mole of magnesium nitride on the reaction with an excess of water gives (a) two moles of ammonia (b) two moles of nitric acid (c) one mole of ammonia (d) one mole of nitric acid

Short Answer

Expert verified
The reaction produces two moles of ammonia.

Step by step solution

01

Write the Chemical Equation

The reaction of magnesium nitride (\( ext{Mg}_3 ext{N}_2 \)) with water (\( ext{H}_2 ext{O} \)) produces magnesium hydroxide (\( ext{Mg(OH)}_2 \)) and ammonia (\( ext{NH}_3 \)). The balanced chemical equation is:\[ \text{Mg}_3\text{N}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{Mg(OH)}_2 + 2\text{NH}_3 \]
02

Identify the Product of Interest

The problem is asking us to find the number of moles of ammonia produced when one mole of magnesium nitride reacts with water.
03

Analyze the Balanced Equation

From the balanced chemical equation, we can see that one mole of \( \text{Mg}_3\text{N}_2 \) reacts to produce two moles of \( \text{NH}_3 \).

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Key Concepts

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

Stoichiometry
Stoichiometry is a core concept in chemistry that involves the quantitative relationships between reactants and products in a chemical reaction. It helps you determine how much product you can obtain from a given amount of reactant or how much reactant you need to produce a desired amount of product. When dealing with stoichiometry, it's vital to work with a balanced chemical equation, ensuring that the number of atoms for each element is equal on both sides of the equation. This balance adheres to the law of conservation of mass.

For example, in the reaction between magnesium nitride (\(\text{Mg}_3\text{N}_2\)) and water, our focus is on finding out the moles of ammonia (\(\text{NH}_3\)) formed. The balanced chemical equation shows the stoichiometric relationship, which tells us how many moles of each substance react to form the products. In this specific reaction:
  • 1 mole of magnesium nitride reacts with 6 moles of water.
  • It produces 3 moles of magnesium hydroxide.
  • 2 moles of ammonia are produced.
Thus, stoichiometry guided by a balanced equation provides a clear path to answer questions about quantities involved in chemical reactions.
Balanced Chemical Equation
A balanced chemical equation is crucial in chemistry because it shows the exact proportions of reactants and products in a reaction. To balance an equation, adjust the coefficients (the numbers in front of the chemical formulas) so that there are equal numbers of each type of atom on both sides of the equation, respecting the law of conservation of mass.

In our example, the reaction between magnesium nitride and water is given by the balanced equation:\[\text{Mg}_3\text{N}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{Mg(OH)}_2 + 2\text{NH}_3\]
  • Every atom is accounted for, ensuring that atoms are conserved during the reaction.
  • Balancing is achieved by ensuring the smallest possible whole number coefficients.
This equation tells us that 1 mole of magnesium nitride requires 6 moles of water to produce 2 moles of ammonia and 3 moles of magnesium hydroxide. Without this balance, it would be challenging to understand the quantities required or produced.
Moles Concept
The mole is a basic unit in chemistry used to express the amount of a chemical substance. It is a bridge between the atomic world and the macroscopic world we interact with daily. One mole contains exactly \(6.022 \times 10^{23}\) particles of the substance (Avogadro's number), whether they are atoms, molecules, or ions.

In the context of the given exercise, the concept of moles allows us to relate masses of substances to the number of particles and reactions between them. When considering the reaction of magnesium nitride with water, the balanced equation:\[\text{Mg}_3\text{N}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{Mg(OH)}_2 + 2\text{NH}_3\]facilitates calculations using moles. When one mole of magnesium nitride reacts, it generates two moles of ammonia.

  • The mole concept converts macroscopic measurements (grams, liters) to micro-scale counts (number of molecules or atoms).
  • It allows for accurate scaling of reactions when given masses or volumes in practical applications.
This foundational concept is essential for calculating how much reactant is needed or product is formed in any chemical reaction.

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Most popular questions from this chapter

For decolourization of one mole of \(\mathrm{KMnO}_{4}\) the number of moles of \(\mathrm{H}_{2} \mathrm{O}_{2}\) required are (a) \(1 / 2\) (b) \(3 / 2\) (c) \(5 / 2\) (d) \(7 / 2\)

Consider the following reactions: (1) \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{HX} \longrightarrow \mathrm{X}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (2) \(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{O}_{3} \longrightarrow 2 \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O}\) Here, hydrogen peroxide acts as (a) an oxidizing agent in both (1) and (2) (b) an oxidizing agent in (2) and reducing agent in (1) (c) an oxidizing agent in (1) and a reducing agent in (2) (d) a reducing agent in both (1) and (2)

Identify the two liquids \((\mathrm{X})\) and \((\mathrm{Y})\) made up of same atoms, both \((\mathrm{X})\) and \((\mathrm{Y})\) are diamagnetic. (X) turns blue litmus to red, but (Y) does not. Also \(\mathrm{X}+\mathrm{KI}+\mathrm{H}^{+} \stackrel{\text { Starch }}{\longrightarrow}\) Blue colour \(\mathrm{Y}+\mathrm{KI}+\mathrm{H}^{+} \stackrel{\text { Starch }}{\longrightarrow}\) No colour (a) \(\mathrm{X}=\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Y}=\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{X}=\mathrm{H}_{2} \mathrm{O}, \mathrm{Y}=\mathrm{H}_{2} \mathrm{O}_{2}\) (c) Both \(\mathrm{X}\) and \(\mathrm{Y}\) are \(\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{X}=\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Y}=\mathrm{O}_{3}\)

Ammonium compound which does not give \(\mathrm{NH}_{3}\) on heating is (a) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) (b) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) (d) \(\mathrm{NH}_{4} \mathrm{Cl}\)

Match the following Column-I (Manufacturing process) (a) Deacon's process for chlorine (b) Hydrogenation of vegetable oils (c) Ostwald's process for nitric acid (d) Haber's process for ammonia Column-II (Catalyst used)] (p) Finely divided iron with molybdenum as promoter (q) Copper (II) chloride (r) Finely divided nickel powder (s) Platinum gauze

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