Question

The last step in the production of nitric acid is the reaction of nitrogen dioxide with water. $$\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$$ (a) Balance the chemical equation. (b) If \(65.0 \mathrm{~g}\) of nitrogen dioxide is reacted with excess water, calculate the theoretical yield. (c) If only \(43.8 \mathrm{~g}\) of nitric acid is obtained, calculate the percent yield.

Short answer

(a) 2NO₂ + H₂O → 2HNO₃ + NO; (b) 89.03 g HNO₃; (c) 49.2% yield.

Step-by-step solution

Write Unbalanced Equation

Start with the given unbalanced chemical equation: \( \mathrm{NO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{HNO}_{3}(aq) + \mathrm{NO}(g) \). This needs to be balanced first.

Balance the Chemical Equation

Balance the chemical equation by ensuring the number of atoms for each element is equal on both sides. The balanced equation is: 2NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g).

Calculate Molar Masses

Calculate the molar mass of NO₂ (46.01 g/mol) and HNO₃ (63.01 g/mol) using the atomic masses: N=14.01, O=16.00, H=1.01.

Convert Grams of NO₂ to Moles

Convert 65.0 g of NO₂ to moles using its molar mass: \( \frac{65.0 \text{ g}}{46.01 \text{ g/mol}} \approx 1.413 \text{ moles of NO}_2 \).

Use Stoichiometry to Find Moles of HNO₃

According to the balanced equation, 2 moles of NO₂ produce 2 moles of HNO₃. Therefore, 1.413 moles of NO₂ will produce 1.413 moles of HNO₃.

Calculate Theoretical Yield in Grams

Convert moles of HNO₃ to grams: \( 1.413 \text{ moles} \times 63.01 \text{ g/mol} = 89.03 \text{ g of HNO}_3 \). This is the theoretical yield of HNO₃.

Calculate Percent Yield

Percent yield is calculated using the formula: \( \left(\frac{\text{actual yield}}{\text{theoretical yield}}\right) \times 100\% \). Therefore, \( \left(\frac{43.8 \text{ g}}{89.03 \text{ g}}\right) \times 100\% \approx 49.2\% \).

Key concepts

Stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions, allowing us to predict how much of a product will be formed from given amounts of reactants. It involves the use of balanced chemical equations to determine the proportions of reactants and products involved in a chemical reaction. In our nitric acid production exercise, stoichiometry is employed to calculate the theoretical yield of nitric acid from nitrogen dioxide.
The process begins with a balanced chemical equation, which provides the essential ratios of reactants and products. For example, in the equation: \( 2\text{NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow 2\text{HNO}_3(aq) + \text{NO}(g) \), the coefficients indicate that two moles of nitrogen dioxide react with one mole of water to produce two moles of nitric acid and one mole of nitric oxide.

• The molar masses of the substances are essential. They allow the conversion of mass to moles, which is the basis of stoichiometric calculations.
• Stoichiometry helps in determining how much of a reactant is needed and how much product will be created.

In this exercise, using stoichiometry, 1.413 moles of \( \text{NO}_2 \) (determined from the given mass and its molar mass) directly correlates to 1.413 moles of \( \text{HNO}_3 \), showcasing the 1:1 molar ratio in our balanced equation.

Balancing Chemical Equations

Balancing chemical equations is crucial because it ensures the law of conservation of mass is obeyed, meaning matter is neither created nor destroyed during a chemical reaction. When we balance a chemical equation, we make sure that the number of atoms for each element is the same on both sides of the equation.
Let's break down the balancing process for the nitric acid production equation:

• Start with the unbalanced equation: \( \text{NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow \text{HNO}_3(aq) + \text{NO}(g) \).
• Identify the number of each type of atom on both sides of the equation. Notice that nitrogen and oxygen atoms must be balanced.
• Balance the nitrogen atoms by ensuring that there are 2 moles of \( \text{NO}_2 \) and 2 moles of \( \text{HNO}_3 \), creating an equal mole of \( \text{NO} \) gas.
• Now ensure that oxygen and hydrogen are balanced. In this case, they are balanced in accordance with the nitrogen molecules.
• The balanced equation is: \( 2\text{NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow 2\text{HNO}_3(aq) + \text{NO}(g) \).

This final balanced equation confirms that the law of conservation of mass is honored, as it accounts for every atom present before and after the reaction.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, comparing the actual yield of a product to the theoretical yield.
The theoretical yield is the maximum amount of product that could be formed from given reactants, as calculated through stoichiometry, assuming all reactants are converted with no loss or error.

• The actual yield is the amount of product actually recovered from an experimental procedure. It is often less due to various practical limitations such as incomplete reactions or product loss.

Calculating the percent yield involves using the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \].
In our example, the actual yield of nitric acid was found to be 43.8 grams, and the theoretical yield was calculated as 89.03 grams. Applying the percent yield formula yields \((43.8 / 89.03) \times 100 \approx 49.2\%\). This indicates that slightly less than half of the theoretical nitric acid was obtained, showing the efficiency of the reaction was about 49.2%, likely due to experimental imperfections.