Problem 67 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

Balanced equations are: (a) already balanced, (b) Step 2: \( \mathrm{UO}_{2} + 4\mathrm{HF} \rightarrow \mathrm{UF}_{4} + 2\mathrm{H}_{2} \mathrm{O} \), \( \mathrm{UF}_{4} + \mathrm{F}_{2} \rightarrow \mathrm{UF}_{6} \), (c) Step 4: \( \mathrm{TiO}_{2} + 2\mathrm{Cl}_{2} + 2\mathrm{C} \rightarrow \mathrm{TiCl}_{4} + 2\mathrm{CO} \) and \( \mathrm{TiCl}_{4} + 2\mathrm{Mg} \rightarrow \mathrm{Ti} + 2\mathrm{MgCl}_{2} \).

Step 1: Analyze the Urea Synthesis Equation

We begin with the equation \( \mathrm{CO}_{2} + \mathrm{NH}_{3} \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2} + \mathrm{H}_{2} \mathrm{O} \). Count the atoms of each element on both sides. On the left, we have 1 Carbon (C), 1 Oxygen (O), and 3 Hydrogen (H) from \( \mathrm{NH}_3 \) and 2 from \( \mathrm{CO}_2 \). On the right, we have 1 C, 1 N in \( \mathrm{NH}_2 \mathrm{CONH}_2 \), and 2 H from \( \mathrm{H}_2 \mathrm{O} \) combining with 4 total H from \( \mathrm{NH}_2 \mathrm{CONH}_2 \). The equation is thus already balanced.

Step 2: Analyze First Uranium Equation

The equation \( \mathrm{UO}_{2} + \mathrm{HF} \rightarrow \mathrm{UF}_{4} + \mathrm{H}_{2} \mathrm{O} \) must balance. Start with uranium (U) which has 1 atom on both sides. For fluorine (F), add a coefficient of 4 before \( \mathrm{HF} \) to get 4 F atoms. Oxygen (O) and hydrogen (H) atoms are already balanced with one \( \mathrm{H}_{2} \mathrm{O} \). Thus, the balanced equation is \( \mathrm{UO}_{2} + 4\mathrm{HF} \rightarrow \mathrm{UF}_{4} + 2\mathrm{H}_{2} \mathrm{O} \).

Step 3: Analyze Second Uranium Equation

For the equation \( \mathrm{UF}_{4} + \mathrm{F}_{2} \rightarrow \mathrm{UF}_{6} \), balance fluorine by ensuring 2 extra fluorines are added, which needs one molecule of \( \mathrm{F}_{2} \). The equation is balanced as \( \mathrm{UF}_{4} + \mathrm{F}_{2} \rightarrow \mathrm{UF}_{6} \).

Step 4: Analyze First Titanium Equation

Looking at \( \mathrm{TiO}_{2} + \mathrm{Cl}_{2} + \mathrm{C} \rightarrow \mathrm{TiCl}_{4} + \mathrm{CO} \), the titanium (Ti) is already balanced. For chlorine (Cl), adjust Cl2 to 2, and for carbon (C), 2 CO are required. So the equation should have a coefficient of 2 before both \( \mathrm{Cl}_2 \) and \( \mathrm{CO} \): \( \mathrm{TiO}_{2} + 2\mathrm{Cl}_{2} + 2\mathrm{C} \rightarrow \mathrm{TiCl}_{4} + 2\mathrm{CO} \).

Step 5: Analyze Second Titanium Equation

For \( \mathrm{TiCl}_{4} + \mathrm{Mg} \rightarrow \mathrm{Ti} + \mathrm{MgCl}_{2} \), titanium (Ti) is balanced by default. One needs two \( \mathrm{Mg} \) atoms to balance 2 \( \mathrm{Cl}_{2} \). The equation becomes \( \mathrm{TiCl}_{4} + 2\mathrm{Mg} \rightarrow \mathrm{Ti} + 2\mathrm{MgCl}_{2} \).

Key Concepts

Urea Synthesis

Urea synthesis is a crucial process in the chemical industry, primarily for the production of fertilizers. The chemical reaction can be represented by the equation: \[\mathrm{CO}_{2} + \mathrm{2NH}_{3} \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2} + \mathrm{H}_{2} \mathrm{O}.\]This process involves the combination of carbon dioxide \(\mathrm{CO}_2\) and ammonia \(\mathrm{NH}_3\). By balancing the equation, the number of atoms for each element should be the same on both sides. Here are some key points about the reaction:

• Ammonia acts as both a nucleophile and a source of nitrogen.
• The reaction results in the formation of urea \(\mathrm{NH}_{2} \mathrm{CONH}_{2}\), a widely used nitrogenous fertilizer.
• Water \(\mathrm{H}_2\mathrm{O}\) is a byproduct in this reaction, indicating the necessity of proper balance.

It’s crucial to understand that each reactant plays a significant role in determining the output of the reaction. A clear chemical equation helps in understanding the stoichiometry, which is vital for practical applications in industries.
In industries, this reaction does not occur in a single step but via intermediates, which requires control over the temperature and pressure to ensure maximum yield.

Uranium Enrichment

Uranium enrichment is a critical process for producing nuclear fuel. The reactions involved in uranium enrichment include:

• First, uranium dioxide \(\mathrm{UO}_2\) reacts with hydrofluoric acid \(\mathrm{HF}\) to form uranium tetrafluoride \(\mathrm{UF}_4\) and water \(\mathrm{H}_2\mathrm{O}\).
• Then, \(\mathrm{UF}_4\) reacts with fluorine gas \(\mathrm{F}_2\) to produce uranium hexafluoride \(\mathrm{UF}_6\).

These reactions are represented by:

• \(\mathrm{UO}_{2} + 4 \mathrm{HF} \rightarrow \mathrm{UF}_{4} + 2\mathrm{H}_{2} \mathrm{O}\)
• \(\mathrm{UF}_{4} + \mathrm{F}_{2} \rightarrow \mathrm{UF}_{6}\)

Understanding these reactions is crucial because:

• Uranium hexafluoride \(\mathrm{UF}_6\) is a vital intermediate in uranium enrichment, where it is further processed to separate different isotopes of uranium.
• Balancing such equations is essential for accurate prediction of required reactants and produced byproducts.
• The process requires careful handling as uranium compounds and fluorine gases are highly reactive and toxic.

Thus, uranium enrichment involves precise chemical processes to ensure the safe and efficient production of enriched uranium suitable for fuel in nuclear power plants.

Titanium Production

Titanium production involves complex chemical reactions to extract titanium metal from its ore. Two main reactions are used in this process:

• The first reaction is the conversion of titanium dioxide \(\mathrm{TiO}_2\) into titanium tetrachloride \(\mathrm{TiCl}_4\) using chlorine gas \(\mathrm{Cl}_2\) and carbon \(\mathrm{C}\).
• The second is the reduction of titanium tetrachloride to titanium metal using magnesium \(\mathrm{Mg}\).

The equations for these reactions are:

• \(\mathrm{TiO}_{2} + 2\mathrm{Cl}_{2} + 2\mathrm{C} \rightarrow \mathrm{TiCl}_{4} + 2\mathrm{CO}\)
• \(\mathrm{TiCl}_{4} + 2\mathrm{Mg} \rightarrow \mathrm{Ti} + 2\mathrm{MgCl}_{2}\)

These reactions help to understand the following aspects:

• The first reaction forms titanium tetrachloride, a volatile compound easily purified by distillation.
• The second reaction, known as the Kroll process, reduces \(\mathrm{TiCl}_4\) to metallic titanium.
• Balancing these equations is crucial for industrial optimization, ensuring that no excess reactants are wasted.

The production of titanium illustrates the importance of chemical equation balancing in maximizing yield and efficiency. Simple stoichiometric understanding aids in scaling up these reactions for industrial applications, meeting global titanium demands.