Problem 45 - Chemistry and Chemical Reactivity

Question

Balance the following equations: (a) The synthesis of urea, a common fertilizer \(\mathrm{CO}_{2}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \longrightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (b) Reactions used to make uranium(VI) fluoride for the enrichment of natural uranium \(\mathrm{UO}_{2}(\mathrm{s})+\mathrm{HF}(\mathrm{aq}) \longrightarrow \mathrm{UF}_{4}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\) \(\mathrm{UF}_{4}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{g}) \longrightarrow \mathrm{UF}_{6}(\mathrm{s})\) (c) The reaction to make titanium(IV) chloride, which is then converted to titanium metal \(\mathrm{TiO}_{2}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{TiCl}_{4}(\ell)+\mathrm{CO}(\mathrm{g})\) $$ \mathrm{TiCl}_{4}(\ell)+\mathrm{Mg}(\mathrm{s}) \longrightarrow \mathrm{Ti}(\mathrm{s})+\mathrm{MgCl}_{2}(\mathrm{s}) $$

Step-by-Step Solution

Answer

Balanced equations: (a) \( \mathrm{CO}_{2} + 2\mathrm{NH}_{3} \rightarrow \mathrm{NH}_{2}\mathrm{CONH}_{2} + \mathrm{H}_{2} \mathrm{O} \) (b) \( \mathrm{UO}_{2} + 4\mathrm{HF} \rightarrow \mathrm{UF}_{4} + 2\mathrm{H}_{2}\mathrm{O} \), \( \mathrm{UF}_{4} + \mathrm{F}_{2} \rightarrow \mathrm{UF}_{6} \) (c) \( \mathrm{TiO}_{2} + 2\mathrm{Cl}_{2} + \mathrm{C} \rightarrow \mathrm{TiCl}_{4} + 2\mathrm{CO} \), \( \mathrm{TiCl}_{4} + 2\mathrm{Mg} \rightarrow \mathrm{Ti} + 2\mathrm{MgCl}_{2} \)

Step 1: Analyze the First Reaction (a)

The chemical equation for the synthesis of urea is given as \( \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell) \). Start by counting the number of atoms of each element on both sides of the equation. We have:- Reactants: C = 1, O = 2, N = 1, H = 3 - Products: C = 1, O = 1 + 1 (from H\(_2\)O), N = 2, H = 4 (2 from NH\(_2\)CONH\(_2\) + 2 from H\(_2\)O)Notice that there is an imbalance. Specifically, the reactants have one nitrogen while the products have two. Also, we need more hydrogens in the reactants to match the products. This means we need to use two molecules of ammonia (NH\(_3\)):

Step 2: Balance the First Reaction (a)

Adjust the equation by putting a coefficient of 2 in front of \( \mathrm{NH}_{3}(\mathrm{g}) \):\[\mathrm{CO}_{2}(\mathrm{g})+2\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\]With this change:- Reactants: C = 1, O = 2, N = 2, H = 6- Products: C = 1, O = 2, N = 2, H = 6The equation is now balanced.

Step 3: Analyze the Second Reaction (b, Equation 1)

Consider the reaction:\( \mathrm{UO}_{2}(\mathrm{s})+\mathrm{HF}(\mathrm{aq}) \rightarrow \mathrm{UF}_{4}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell) \). Count the atoms:- Reactants: U = 1, O = 2, H = 1, F = 1- Products: U = 1, F = 4, H = 2, O = 1Add coefficients to balance the equation. Uranium is already balanced but fluorine on the right requires four atoms. Add 4 in front of HF:

Step 4: Balance the Second Reaction (b, Equation 1)

Balance the equation:\[\mathrm{UO}_{2}(\mathrm{s})+4\mathrm{HF}(\mathrm{aq}) \rightarrow \mathrm{UF}_{4}(\mathrm{s})+2\mathrm{H}_{2} \mathrm{O}(\ell)\]Now, the atom counts are:- Reactants: U = 1, O = 2, H = 4, F = 4- Products: U = 1, F = 4, H = 4, O = 2The equation is balanced.

Step 5: Balance the Third Reaction (b, Equation 2)

Consider the reaction:\( \mathrm{UF}_{4}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{g}) \rightarrow \mathrm{UF}_{6}(\mathrm{s}) \). Count the atoms:- Reactants: U = 1, F = 6- Products: U = 1, F = 6The equation is already balanced with one molecule of \( \mathrm{F}_{2} \) providing the two additional fluorines needed.

Step 6: Analyze the Fourth Reaction (c, Equation 1)

Consider the reaction:\( \mathrm{TiO}_{2}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{TiCl}_{4}(\ell)+\mathrm{CO}(\mathrm{g}) \). Count the atoms:- Reactants: Ti = 1, O = 2, Cl = 2, C = 1- Products: Ti = 1, Cl = 4, C = 1, O = 1Notice that Titanium and Carbon are balanced, but Chlorine and Oxygen are not. Add coefficients to balance these.

Step 7: Balance the Fourth Reaction (c, Equation 1)

Introduce a coefficient of 2 in front of \( \mathrm{Cl}_{2} \) and \( \mathrm{CO} \). Adjust the number of \( \mathrm{Cl}_{2} \) and \( \mathrm{CO} \):\[\mathrm{TiO}_{2}(\mathrm{s})+2\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{TiCl}_{4}(\ell)+2\mathrm{CO}(\mathrm{g})\]Ensure balance:- Reactants: Ti = 1, Cl = 4, C = 1, O = 2- Products: Ti = 1, Cl = 4, C = 2, O = 2The equation is now balanced for all elements.

Step 8: Balance the Fifth Reaction (c, Equation 2)

Consider the reaction:\( \mathrm{TiCl}_{4}(\ell)+\mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Ti}(\mathrm{s})+\mathrm{MgCl}_{2}(\mathrm{s}) \). Count the atoms:- Reactants: Ti = 1, Cl = 4, Mg = 1- Products: Ti = 1, Cl = 2, Mg = 1To balance, add a coefficient of 2 in front of \( \mathrm{Mg} \) and \( \mathrm{MgCl}_{2} \):\[\mathrm{TiCl}_{4}(\ell)+2\mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Ti}(\mathrm{s})+2\mathrm{MgCl}_{2}(\mathrm{s})\]Now, the counts are equal for every element.

Key Concepts

Synthesis of Urea

The synthesis of urea is an important chemical process widely used in agriculture as a source of the nitrogen required by plants. Urea is formed from the reaction between carbon dioxide (CO extsubscript{2}) and ammonia (NH extsubscript{3}). The balanced chemical equation for this reaction is:

• Reactants:
  • 1 molecule of carbon dioxide
  • 2 molecules of ammonia
• Products:
  • 1 molecule of urea ( ext{NH extsubscript{2}CONH extsubscript{2}})
  • 1 molecule of water

Balancing this equation involves ensuring the same number of each type of atom on both sides. Urea contains two nitrogen atoms, so two molecules of ammonia are needed to provide the required nitrogen. This equation highlights the principle of stoichiometry, ensuring that all atoms in the reactants are accounted for in the products. The synthesis of urea exemplifies a critical application of chemical equation balancing in industrial chemistry, helping efficiently produce this crucial fertilizer.

Uranium Enrichment

Uranium enrichment is a key process in the production and preparation of nuclear fuel. It involves a series of chemical reactions that ultimately provide the highly sought-after uranium-235 isotope. One of the processes involved is the conversion of uranium dioxide (UO extsubscript{2}) to uranium hexafluoride (UF extsubscript{6}). Here's how it starts:

• Reactants:
  • 1 molecule of uranium dioxide (UO extsubscript{2})
  • 4 molecules of hydrogen fluoride (HF)
• Products:
  • 1 molecule of uranium tetrafluoride (UF extsubscript{4})
  • 2 molecules of water

This balanced equation shows the introduction of fluoride to convert uranium into a form that can be further processed. Step two in this enrichment involves fluorination of UF extsubscript{4} to produce UF extsubscript{6}, another balanced step in sequence:

• Reactants:
  • 1 molecule of UF extsubscript{4}
  • 1 molecule of fluorine gas (F extsubscript{2})
• Products:
  • 1 molecule of uranium hexafluoride (UF extsubscript{6})

Each stage involves careful balance to comply with chemical principles and ensure yield optimization in nuclear energy applications.

Titanium Production

Titanium production involves multiple steps to convert titanium ores into usable metal. A notable step is the conversion of titanium dioxide (TiO extsubscript{2}) to titanium tetrachloride (TiCl extsubscript{4}). We start by chlorination:

• Reactants:
  • 1 molecule of TiO extsubscript{2}
  • 2 molecules of chlorine gas (Cl extsubscript{2})
  • 1 molecule of carbon
• Products:
  • 1 molecule of TiCl extsubscript{4}
  • 2 molecules of carbon monoxide

Balancing this reaction is crucial as it ensures the efficient transformation of raw materials. The next step is reducing TiCl extsubscript{4} to titanium metal using magnesium:

• Reactants:
  • 1 molecule of TiCl extsubscript{4}
  • 2 atoms of magnesium
• Products:
  • 1 atom of titanium
  • 2 molecules of magnesium chloride (MgCl extsubscript{2})

This sequence is vital for producing pure titanium metal, a material extensively used for its durability and strength in various industries.

Stoichiometry

Stoichiometry is a foundational concept in chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. The principle ensures that equations conform to the law of conservation of mass, meaning no atoms are lost or gained but rearranged. When balancing equations, stoichiometry allows us to determine the exact amounts of each reactant needed to produce a desired amount of product.

• Stoichiometric Coefficients: Indicate proportions of reactants and products in a balanced equation. For example, in the synthesis of urea, the coefficient of 2 placed before NH extsubscript{3} ensures nitrogen atoms are balanced.
• Mole Ratios: Used to convert between amounts of reactants or products using their coefficients, facilitating conversions essential for larger-scale chemical production.
• Limiting Reactant: The reactant that determines the maximum amount of product that can be formed, highlighted in synthesis and enrichment processes.

Understanding these components is crucial for accurately predicting reaction outcomes and optimizing the use of resources in industrial applications. Whether creating fertilizers or processing nuclear materials, stoichiometry plays a vital role in achieving desired results efficiently.