Problem 80 - Chemistry The Central Science In Si Units

Question

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to NO: $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ In a certain experiment, \(2.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.50 \mathrm{~g}\) of \(\mathrm{O}_{2} .\) (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and \(\mathrm{H}_{2} \mathrm{O}\) form? \((\mathbf{c})\) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Step-by-Step Solution

Answer

The limiting reactant is \(\mathrm{O}_2\). Forms: \(1.88\,\text{g of } \mathrm{NO}\), \(1.69\,\text{g of } \mathrm{H}_2O\). \(0.928\,\text{g of } \mathrm{NH}_3\) remains.

Step 1: Calculate Moles of Reactants

First, find the molar masses of the reactants. The molar mass of \(\mathrm{NH}_3\) is approximately \(14.01 + 3 \times 1.01 = 17.03\,\text{g/mol}\). The molar mass of \(\mathrm{O}_2\) is approximately \(2 \times 16.00 = 32.00\,\text{g/mol}\). Now, calculate the moles of each reactant: \[\text{Moles of } \mathrm{NH}_3 = \frac{2.00\,\text{g}}{17.03\,\text{g/mol}} \approx 0.117\,\text{mol}\]\[\text{Moles of } \mathrm{O}_2 = \frac{2.50\,\text{g}}{32.00\,\text{g/mol}} \approx 0.0781\,\text{mol}\]

Step 2: Determine Limiting Reactant

The balanced equation is \[ 4 \mathrm{NH}_3(g) + 5 \mathrm{O}_2(g) \rightarrow 4 \mathrm{NO}(g) + 6 \mathrm{H}_2O(g) \]Calculate the mole ratio for \(\mathrm{NH}_3\) and \(\mathrm{O}_2\) using their coefficients: - \(4\) moles of \(\mathrm{NH}_3\) react with \(5\) moles of \(\mathrm{O}_2\).This means for 0.117 moles \(\mathrm{NH}_3\), we need \[ \text{Required moles of } \mathrm{O}_2 = \frac{5}{4} \times 0.117 = 0.146\,\text{mol of } \mathrm{O}_2 \]Since there are only 0.0781 moles of \(\mathrm{O}_2\) available, \(\mathrm{O}_2\) is the limiting reactant.

Step 3: Calculate Grams of Products Formed

Since \(\mathrm{O}_2\) is the limiting reactant, the amount of products formed is determined by the amount of \(\mathrm{O}_2\) present. Using the stoichiometry from the balanced equation:- \(5\) moles of \(\mathrm{O}_2\) produces \(4\) moles of \(\mathrm{NO}\) and \(6\) moles of \(\mathrm{H}_2O\).For \(0.0781\,\text{mol of } \mathrm{O}_2\):\[\text{Moles of } \mathrm{NO} = \frac{4}{5} \times 0.0781 = 0.0625\,\text{mol}\]\[\text{Moles of } \mathrm{H}_2O = \frac{6}{5} \times 0.0781 = 0.0937\,\text{mol}\]Convert these moles to grams:\[\text{Mass of } \mathrm{NO} = 0.0625\,\text{mol} \times 30.01\,\text{g/mol} = 1.88\,\text{g}\]\[\text{Mass of } \mathrm{H}_2O = 0.0937\,\text{mol} \times 18.02\,\text{g/mol} = 1.69\,\text{g}\]

Step 4: Calculate Excess Reactant Remaining

We know \(\mathrm{NH}_3\) is the excess reactant. Use the mole requirement from the limiting reactant to find consumption:\[\text{Required moles of } \mathrm{NH}_3 = \frac{4}{5} \times 0.0781 = 0.0625\,\text{mol}\]\[\text{Used } \mathrm{NH}_3 = 0.117\,\text{mol (initial)} - 0.0625\,\text{mol (consumed)} = 0.0545\,\text{mol}\]Convert to grams:\[\text{Mass of excess } \mathrm{NH}_3 = 0.0545\,\text{mol} \times 17.03\,\text{g/mol} = 0.928\,\text{g}\]

Step 5: Verify Conservation of Mass

Start with the total mass of reactants:\[2.00\,\text{g of } \mathrm{NH}_3 + 2.50\,\text{g of } \mathrm{O}_2 = 4.50\,\text{g}\]The total mass of products and remaining excess:\[1.88\,\text{g of } \mathrm{NO} + 1.69\,\text{g of } \mathrm{H}_2O + 0.928\,\text{g remaining } \mathrm{NH}_3 = 4.498\,\text{g}\]The mass conservation balances within experimental error, verifying the calculations.

Key Concepts

Stoichiometry

Stoichiometry is a key concept in chemistry that focuses on the quantitative analysis of substances involved in chemical reactions. It involves the calculation of reactants and products using balanced chemical equations. The central idea is that the coefficients in a balanced equation convey the molar ratios of the reactants and products.

In the given exercise, stoichiometry is applied to determine how much of each reactant is needed and what amounts of products are formed. By calculating the moles of ammonia (\(\text{NH}_3\)) and oxygen (\(\text{O}_2\)), we can establish their mole ratio through the following steps:

• Find the moles of reactants using their mass and molar mass.
• Use the balanced equation \(4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O}\) to determine the stoichiometric ratio.
• Identify the limiting reactant based on the amount of reactants available.

Understanding stoichiometry allows us to predict the amounts of products based on the limited amount of reactants, which is crucial for efficiently conducting chemical reactions.

Chemical Reaction

A chemical reaction is a process where substances called reactants are transformed into different substances, or products. In any reaction, bonds between atoms are broken and new ones are formed, converting reactants to products while maintaining the same overall mass and number of atoms.

The reaction in the exercise is the conversion of ammonia to nitric oxide with oxygen:\[4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(g)\]. Recognizing that this equation is balanced is critical for understanding which substances are involved and in what proportions. It guides through:

• The determination of how much of each reactant is needed.
• The calculation of excess and limiting reactants.
• The prediction of the masses of the products formed.

Practicing chemical reactions involves understanding how reactants interact and rearrange to form new substances, which is foundational knowledge in chemistry.

Conservation of Mass

The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. This principle means that the mass of the reactants must equal the mass of the products in a closed system.

In the exercise, we verify this law by calculating the total mass of reactants and comparing it with the total mass of products and any excess reactant. The mass of reactants is calculated as follows:

• \(\text{NH}_3\) mass: 2.00 g
• \(\text{O}_2\) mass: 2.50 g

This gives a total initial mass of 4.50 g. Upon completion of the reaction, we calculate the mass of the products:\(1.88 \text{g of NO} + 1.69 \text{g of } \text{H}_2\text{O} + 0.928 \text{g of remaining } \text{NH}_3\). The sum is approximately 4.498 g, which closely matches the initial mass, considering minor experimental errors. This step ensures that our stoichiometric and chemical calculations are grounded in this fundamental law of chemistry.