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Balancing chemical equations is a crucial step in solving stoichiometry problems. The goal is to ensure that the number of atoms for each element is the same on both sides of the equation. This is because matter cannot be created or destroyed in a chemical reaction. In our reaction, the unbalanced equation was:

• \( \text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{(NH}_4)_2\text{SO}_4 \)

Initially, this equation is not balanced because the number of nitrogen and hydrogen atoms aren't the same on both sides. To balance it, we need 2 molecules of \(\text{NH}_3\) reacting with one molecule of \(\text{H}_2\text{SO}_4\). This makes the balanced equation:

• \(2 \text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{(NH}_4)_2\text{SO}_4 \)

Now, the count for each type of atom matches on both sides of the equation, confirming it is balanced.

The Ideal Gas Law is an essential equation in chemistry that relates the pressure, volume, temperature, and amount of gas. It is given by:

• \( PV = nRT \)

Where:

• \(P\) = Pressure in atmospheres (atm)
• \(V\) = Volume in liters (L)
• \(n\) = Moles of gas
• \(R\) = Ideal gas constant (0.0821 L atm/mol K)
• \(T\) = Temperature in Kelvin (K)

In the problem, we used the Ideal Gas Law to find the volume of \(\text{NH}_3\) needed. We have:

• \(P = 1.15\) atm
• \(n = 2.270\) mol
• \(T = 15^{\circ}\text{C} = 288.15 \text{ K}\)

Solving for \(V\), we substituted these values into the equation:\[ V = \frac{nRT}{P} = \frac{2.270 \times 0.0821 \times 288.15}{1.15} \approx 46.39 \, \text{L} \]So, the volume of \(\text{NH}_3\) required is approximately \(46.39\) liters.

Calculating the molar mass of a compound allows us to understand how many grams one mole of that compound weighs. This is essential when determining moles from a given mass in stoichiometry. For ammonium sulfate, the molar mass is calculated as follows:

• Ammonium ion \((\text{NH}_4^+)\): \(14.01 + 4 \times 1.01 = 18.05\)
• There are two ammonium ions, so: \(2 \times 18.05\)
• Sulfur atom \((\text{S})\): \(32.07\)
• Four oxygen atoms \((4 \times 16.00)\)

Adding these together gives the molar mass of ammonium sulfate \(\text{(NH}_4)_2\text{SO}_4\):

• \(2(18.05) + 32.07 + 64.00 = 132.14 \, \text{g/mol}\)

This molar mass is used to convert the given mass of ammonium sulfate in the problem to moles.

Ammonium sulfate \((\text{(NH}_4)_2\text{SO}_4)\) is a chemical compound widely used in fertilizers as a source of nitrogen and sulfur for plants. It is a crystalline salt that dissolves easily in water, making it useful not only in agriculture but also in various industrial applications.To produce ammonium sulfate, ammonia \((\text{NH}_3)\) reacts with sulfuric acid \((\text{H}_2\text{SO}_4)\), forming the fertilizer in a balanced chemical reaction. The resulting compound helps in soil enrichment by providing key nutrients. Moreover, its relatively low cost and ease of production contribute to its widespread use in farming.Understanding the reaction and applications of ammonium sulfate is crucial in fields like agriculture and chemistry where improving crop yield and fertilization efficiency are prioritized.