Question

Magnesium burns in air to produce magnesium oxide, \(\mathrm{MgO},\) and magnesium nitride, \(\mathrm{Mg}_{3} \mathrm{~N}_{2} .\) Magnesium nitride reacts with water to give ammonia. $$ \mathrm{Mg}_{3} \mathrm{~N}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{NH}_{3}(g) $$ What volume of ammonia gas at \(24^{\circ} \mathrm{C}\) and \(753 \mathrm{mmHg}\) will be produced from \(3.93 \mathrm{~g}\) of magnesium nitride?

Short answer

1.90 L of ammonia gas will be produced.

Step-by-step solution

Determine the Molar Mass of Magnesium Nitride

First, calculate the molar mass of magnesium nitride, \(\mathrm{Mg}_3\mathrm{N}_2\). Magnesium has a molar mass of approximately 24.31 \(\mathrm{g/mol}\) and nitrogen has a molar mass of approximately 14.01 \(\mathrm{g/mol}\). The molar mass of magnesium nitride is calculated as follows: \[3 \times 24.31 + 2 \times 14.01 = 100.95 \, \text{g/mol}.\]

Convert Mass to Moles

Convert the given mass of \(\mathrm{Mg}_3\mathrm{N}_2\) to moles using its molar mass. Using the mass-to-mole conversion formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\), compute: \[\frac{3.93 \,\text{g}}{100.95 \, \text{g/mol}} = 0.0389 \, \text{mol}.\]

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, calculate the moles of ammonia produced: \[0.0389 \, \text{mol of Mg}_3\mathrm{N}_2 \times \frac{2 \, \text{mol of NH}_3}{1 \, \text{mol of Mg}_3\mathrm{N}_2} = 0.0778 \, \text{mol of NH}_3.\]

Use the Ideal Gas Law to Find Volume

Use the ideal gas law to calculate the volume of ammonia gas produced. First, convert the temperature to Kelvin: \[24^\circ\mathrm{C} = 24 + 273.15 = 297.15\, \mathrm{K}.\] Convert pressure to atmospheres: \[753\, \mathrm{mmHg} = \frac{753}{760} \approx 0.991\, \mathrm{atm}.\] Use \(PV = nRT\) with \(R = 0.0821 \, \mathrm{L} \cdot \mathrm{atm} / \mathrm{K} \cdot \mathrm{mol}\) to find \(V:\) \[V = \frac{nRT}{P} = \frac{0.0778 \, \text{mol} \times 0.0821 \, \mathrm{L} \cdot \mathrm{atm} / \mathrm{K} \cdot \mathrm{mol} \times 297.15 \, \mathrm{K}}{0.991 \, \mathrm{atm}} \approx 1.90 \, \mathrm{L}.\]

Key concepts

Molar Mass

The concept of molar mass is essential to understanding how to calculate the number of moles of a substance when its mass is known. Molar mass is defined as the mass of one mole of a substance, measured in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in the chemical formula.

In the given problem, we calculated the molar mass of magnesium nitride, \(\mathrm{Mg}_3\mathrm{N}_2\), which consists of magnesium (Mg) and nitrogen (N). The atomic mass of magnesium is approximately 24.31 g/mol, and nitrogen is about 14.01 g/mol. Since there are three magnesium atoms and two nitrogen atoms in the chemical formula, the molar mass is calculated by:

• 3 times the atomic mass of magnesium: \(3 \times 24.31 = 72.93\) g/mol.
• 2 times the atomic mass of nitrogen: \(2 \times 14.01 = 28.02\) g/mol.
• Add the two results: \(72.93 + 28.02 = 100.95\) g/mol.

This value helps us convert the given mass of \(3.93\) g of magnesium nitride into moles, which is crucial for further calculations in solving the problem.

Chemical Reactions

Understanding chemical reactions is vital, especially when it involves deriving products from reactants based on a given equation. In this exercise, a balanced chemical reaction shows how magnesium nitride reacts with water to produce magnesium hydroxide and ammonia gas.

The balanced equation \(\mathrm{Mg}_3 \mathrm{N}_2(s) + 6 \mathrm{H}_2\mathrm{O}(l) \rightarrow 3 \mathrm{Mg} (\mathrm{OH})_2(s) + 2 \mathrm{NH}_3(g)\) indicates that one molecule of magnesium nitride reacts with six molecules of water to produce three molecules of magnesium hydroxide and two molecules of ammonia.

• 1 mole of \(\mathrm{Mg}_3\mathrm{N}_2\) produces 2 moles of \(\mathrm{NH}_3\).
• This stoichiometric relationship allows us to determine the moles of ammonia produced from a known amount of magnesium nitride.

Identifying and using stoichiometric coefficients in a balanced equation is a critical skill when predicting amounts of products formed.

Gas Laws

Gas laws, particularly the Ideal Gas Law, are crucial when dealing with gases under various conditions of temperature and pressure. The Ideal Gas Law is expressed as \(PV = nRT\), where \(P\) is the pressure in atmospheres, \(V\) is the volume in liters, \(n\) is the number of moles, \(R\) is the universal gas constant (0.0821 L·atm/K·mol), and \(T\) is the temperature in Kelvin.

In this problem, after determining the number of moles of ammonia gas produced (0.0778 moles), we use the Ideal Gas Law to find its volume. Here are the steps involved:

• Convert temperature from Celsius to Kelvin: add 273.15 to 24°C, resulting in 297.15 K.
• Convert pressure from mmHg to atm by dividing by the conversion factor 760 mmHg/atm. Thus, 753 mmHg becomes approximately 0.991 atm.
• Plug these values into the Ideal Gas Law equation: \(V = \frac{nRT}{P} = \frac{0.0778 \times 0.0821 \times 297.15}{0.991}\), which simplifies to 1.90 liters.

This formula enables us to determine the volume of ammonia gas under the specified conditions.

Conversion of Units

Converting units is a fundamental skill in solving chemistry problems, particularly those involving different measurement systems. Being able to switch between units for pressure, volume, temperature, and mass is essential to applying formulas correctly.

In the exercise, several unit conversions were necessary:

• Converted temperature from degrees Celsius to Kelvin by adding 273.15. This is required because gas law calculations require an absolute temperature scale.
• Converted pressure from mmHg to atm by dividing by 760 mmHg per atmosphere, aligning pressure units with those used in the Ideal Gas Law.
• Total moles were derived from mass using the calculated molar mass, essential for understanding the amount of substance in the reactant.

Correct unit conversion ensures that the calculations yield accurate, meaningful results, helping predict outcomes in chemical reactions.