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II-35. Lithium nitride reacts with water to produce ammonia and lithium hydroxide according to the equation $$ \mathrm{Li}_{3} \mathrm{~N}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{NH}_{3}(g)+3 \mathrm{LiOH}(a q) $$ Heavy water is water with the isotope deuterium in place of ordinary hydrogen, and its formula is \(\mathrm{D}_{2} \mathrm{O}\). The above reaction can be used to produce heavy ammonia, \(\mathrm{ND}_{3}(g)\), according to the equation $$ \mathrm{Li}_{9} \mathrm{~N}(s)+3 \mathrm{D}_{2} \mathrm{O}(l) \rightarrow \mathrm{ND}_{3}(g)+3 \mathrm{LiOD}(a q) $$ Calculate how many grams of heavy water are required to produce \(200.0\) milligrams of \(\mathrm{ND}_{3}(g) .\) The atomic mass of deuterium is \(2.014\).

Short Answer

Expert verified
Approximately 0.5992 grams of heavy water are required.

Step by step solution

01

Write the Balanced Equation

The balanced equation for the reaction is given:\[\text{Li}_3 \text{N}(s) + 3\text{D}_2 \text{O}(l) \rightarrow \text{ND}_3(g) + 3\text{LiOD}(aq)\]This equation tells us that 3 moles of \(\text{D}_2 \text{O}\) react to produce 1 mole of \(\text{ND}_3\).
02

Convert Mass of ND3 to Moles

We need to find out how many moles of \(\text{ND}_3\) are present in 200.0 mg of \(\text{ND}_3\). First, convert the mass from milligrams to grams: \[200.0 \text{ mg} = 0.2000 \text{ g}\]The molar mass of \(\text{ND}_3\) is calculated as follows: N (14.01) + 3 * D (2.014) = 20.052 g/mol.Now, calculate moles of \(\text{ND}_3\): \[\text{moles of ND}_3 = \frac{0.2000}{20.052} \approx 0.009977 \text{ moles}\]
03

Use Stoichiometry to Find Moles of D2O

According to the balanced equation, 1 mole of \(\text{ND}_3\) is produced by 3 moles of \(\text{D}_2 \text{O}\). Therefore, to find moles of \(\text{D}_2 \text{O}\) needed:\[\text{moles of D}_2 \text{O} = 3 \times 0.009977 \approx 0.029931 \text{ moles}\]
04

Convert Moles of D2O to Grams

The molar mass of \(\text{D}_2 \text{O}\) is calculated by adding the atomic masses: D (2.014) * 2 + O (16.00) = 20.028 g/mol.Convert moles of \(\text{D}_2 \text{O}\) to grams:\[\text{grams of D}_2 \text{O} = 0.029931 \times 20.028 \approx 0.5992 \text{ g}\]
05

Summary of Calculation

In summary, you need approximately 0.5992 grams of heavy water (\(\text{D}_2 \text{O}\)) to produce 200.0 mg of \(\text{ND}_3\).

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Key Concepts

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

Chemical Reactions
A chemical reaction is a process in which substances called reactants are transformed into different substances known as products. In the given problem, lithium nitride reacts with heavy water (deuterated water) to produce heavy ammonia and lithium deuteroxide. This reaction highlights the process where new compounds are formed through the rearrangement of atoms. Key indicators of a chemical reaction include:
  • Change in color
  • Formation of a precipitate or gas
  • Change in temperature
  • Change in chemical composition
In the case of the problem, the change manifests in the formation of new products like heavy ammonia and lithium deuteroxide. Chemical reactions can be exothermic, releasing heat, or endothermic, absorbing heat. Understanding the nature of the reaction is crucial for predicting the behavior of substances during the process.
Molar Mass
The molar mass of a compound is the mass of one mole of its entities, whether atoms, molecules, ions, etc., usually expressed in grams per mole (g/mol). Molar mass is calculated by adding together the atomic masses of all atoms in a chemical formula. For example, in our problem:
  • Molar mass of deuterium (D) is 2.014 g/mol.
  • Molar mass of nitrogen (N) is 14.01 g/mol.
  • Therefore, the molar mass of heavy ammonia (ND euex3) is N (14.01) + 3 * D (2.014) = 20.052 g/mol.
Understanding molar mass is crucial for converting between grams and moles in chemical equations, a step vital for stoichiometry. This conversion allows us to determine how much of a substance is needed or produced in a given reaction. Precise calculations of molar masses allow chemists to use the right amounts of substances to achieve desired reactions without waste.
Balanced Chemical Equation
A balanced chemical equation accurately represents the conservation of mass and charge by showing the same number of each type of atom on both sides of the equation. In the example provided, the reaction:\[\text{Li}_3\text{N}(s) + 3\text{D}_2\text{O}(l) \rightarrow \text{ND}_3(g) + 3\text{LiOD}(aq)\]This indicates that three molecules of heavy water react with one molecule of lithium nitride to form one molecule of heavy ammonia and three molecules of lithium deuteroxide. Each side of the equation must reflect the same number of each type of atom.Steps to balance an equation include:
  • Write the unbalanced equation.
  • Count atoms of each element in reactants and products.
  • Add coefficients to balance each type of atom.
  • Ensure the same number of each atom on both sides.
Balancing equations is fundamental in stoichiometry, the area of chemistry that quantifies reactants and products, ensuring that no mass is lost during the reaction.

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