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Magnesium nitride, \(\mathrm{Mg}_{3} \mathrm{~N}_{2}\), reacts with water to produce magnesium hydroxide and ammonia. How many grams of ammonia can you obtain from \(7.50 \mathrm{~g}\) of magnesium nitride?

Short Answer

Expert verified
You can obtain 2.53 grams of ammonia.

Step by step solution

01

Write the Balanced Chemical Equation

The reaction between magnesium nitride \( \mathrm{Mg}_3 \mathrm{N}_2 \) and water \( \mathrm{H}_2 \mathrm{O} \) can be represented as follows:\[ \mathrm{Mg}_3 \mathrm{N}_2 + 6 \mathrm{H}_2 \mathrm{O} \rightarrow 3 \mathrm{Mg(OH)}_2 + 2 \mathrm{NH}_3 \]This balanced equation shows that one mole of \( \mathrm{Mg}_3 \mathrm{N}_2 \) reacts with six moles of water to produce three moles of magnesium hydroxide \( \mathrm{Mg(OH)}_2 \) and two moles of ammonia \( \mathrm{NH}_3 \).
02

Calculate Molar Masses

Determine the molar masses of magnesium nitride and ammonia. For \( \mathrm{Mg}_3 \mathrm{N}_2 \), the molar mass is \( 3 \times 24.31 + 2 \times 14.01 = 100.95 \, \mathrm{g/mol} \). For \( \mathrm{NH}_3 \), the molar mass is \( 14.01 + 3 \times 1.01 = 17.04 \, \mathrm{g/mol} \).
03

Calculate Moles of Magnesium Nitride

Use the given mass of magnesium nitride to find the number of moles.\[ \text{Moles of } \mathrm{Mg}_3 \mathrm{N}_2 = \frac{7.50 \, \mathrm{g}}{100.95 \, \mathrm{g/mol}} = 0.0743 \, \mathrm{mol} \]
04

Determine Moles of Ammonia Produced

From the balanced equation, 1 mole of \( \mathrm{Mg}_3 \mathrm{N}_2 \) produces 2 moles of \( \mathrm{NH}_3 \). Therefore, 0.0743 moles of \( \mathrm{Mg}_3 \mathrm{N}_2 \) will produce \( 0.0743 \times 2 = 0.1486 \, \mathrm{mol} \) of \( \mathrm{NH}_3 \).
05

Calculate Mass of Ammonia

Convert the moles of ammonia to grams using its molar mass.\[ \text{Mass of } \mathrm{NH}_3 = 0.1486 \, \mathrm{mol} \times 17.04 \, \mathrm{g/mol} = 2.53 \, \mathrm{g} \]

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Key Concepts

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

Chemical Reactions
Chemical reactions are processes where substances, known as reactants, are transformed into different substances called products. This occurs through the breaking and forming of chemical bonds. Understanding chemical reactions allows us to predict the changes that occur in any given chemical process. In the case of magnesium nitride reacting with water, the reactants are magnesium nitride and water, and the products are magnesium hydroxide and ammonia. The key is to comprehend how these reactants interact to form the products.
A clear understanding of chemical reactions helps in anticipating the type and amount of products formed from specified reactants. It is crucial to know the nature of the reactants and how they can transform, which involves changing bonds in a way that results in new products.
Balanced Chemical Equation
A balanced chemical equation is essential in understanding the stoichiometry of a chemical reaction. It shows the relationship between the number of moles of reactants and products in a chemical reaction. Each element has the same number of atoms on both sides of the equation, maintaining conservation of mass.
For example, the balanced chemical equation for the reaction of magnesium nitride with water is:
  • \( \mathrm{Mg}_3 \mathrm{N}_2 + 6 \mathrm{H}_2 \mathrm{O} \rightarrow 3 \mathrm{Mg(OH)}_2 + 2 \mathrm{NH}_3 \)
This equation indicates that one mole of magnesium nitride reacts with six moles of water to produce three moles of magnesium hydroxide and two moles of ammonia.
Balancing chemical equations helps determine the quantities of reactants needed and the expected amounts of products formed.
Molar Mass Calculation
Molar mass is a fundamental concept used to convert between the mass of a substance and the number of moles. To calculate the molar mass of a compound, you add up the atomic masses of all the atoms in its formula. In our example, the molar mass of magnesium nitride \( \mathrm{Mg}_3 \mathrm{N}_2 \) is calculated as:
  • \( 3 \times 24.31 + 2 \times 14.01 = 100.95 \, \mathrm{g/mol} \)
For ammonia \( \mathrm{NH}_3 \), it is:
  • \( 14.01 + 3 \times 1.01 = 17.04 \, \mathrm{g/mol} \)
These calculations are necessary to transform chemical equations into quantities you can work with in actual experiments. Knowing the molar mass allows you to determine how much of a substance is involved in a reaction, a key step in stoichiometry.
Mole-to-Mass Conversion
Mole-to-mass conversion is a vital step in stoichiometry, allowing us to move between the amount of substance in moles and its mass in grams. Once we know the number of moles, we can use the molar mass to find the mass.
From the exercise, calculate moles of magnesium nitride:
  • \( \frac{7.50 \, \mathrm{g}}{100.95 \, \mathrm{g/mol}} = 0.0743 \, \mathrm{mol} \)
Then use the balanced equation to find moles of ammonia:
  • \( 0.0743 \, \mathrm{mol} \times 2 = 0.1486 \, \mathrm{mol} \)
Finally, convert moles of ammonia to grams:
  • \( 0.1486 \, \mathrm{mol} \times 17.04 \, \mathrm{g/mol} = 2.53 \, \mathrm{g} \)
This conversion is crucial for practical applications, such as calculating the yield of reactions or the necessary amounts of reactants.

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Most popular questions from this chapter

A \(50.00\) -mL volume of \(0.4987 \mathrm{M} \mathrm{HCl}\) was added to a \(5.436-\mathrm{g}\) sample of milk of magnesia. This solution was then titrated with \(0.2456 \mathrm{M} \mathrm{NaOH}\). If it required \(39.42 \mathrm{~mL}\) of \(\mathrm{NaOH}\) to reach the endpoint, what was the mass percentage of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in the milk of magnesia?

The amount of sodium hypochlorite in a bleach solution can be determined by using a given volume of bleach to oxidize excess iodide ion to iodine, because the reaction goes to completion. The amount of iodine produced is then determined by titration with sodium thiosulfate, \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\), which is oxidized to sodium tetrathionate, \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6} .\) Potassium iodide was added in excess to \(5.00 \mathrm{~mL}\) of bleach (density = \(1.00 \mathrm{~g} / \mathrm{mL}\) ). This solution, containing the iodine released in the reaction, was titrated with \(0.100 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\). If \(34.6 \mathrm{~mL}\) of sodium thiosulfate was required to reach the endpoint (detected by disappearance of the blue color of the starch- iodine complex), what was the mass percentage of \(\mathrm{NaClO}\) in the bleach?

Describe the bonding (using valence bond theory) of the Group IVA atoms in each of the following: a. \(\mathrm{CCl}_{4}\) b. HCN c. \(\mathrm{SiF}_{4}\) d. \(\mathrm{CH}_{3} \mathrm{COOH}\) (acetic acid)

Write balanced equations for each of the following. a. Bromine reacts with aqueous sodium hydroxide to give hypobromite and bromide ions. b. Hydrogen bromide gas forms when sodium bromide is heated with phosphoric acid.

Give equations for the preparation of \(\mathrm{N}_{2} \mathrm{O}\). You can use \(\mathrm{NaOH}, \mathrm{NaNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4}\), and \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\left(\right.\) plus \(\mathrm{H}_{2} \mathrm{O}\) ). Sev- eral steps are required.

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