Problem 78 - Chemistry The Central Science

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 of \(\mathrm{H}_{2} \mathrm{O}\) form? (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

(a) The limiting reactant is O₂. (b) 1.875 g of NO and 1.689 g of H₂O are formed. (c) 0.935 g of NH₃ remains. (d) Our calculations are consistent with the law of conservation of mass, as the total mass of reactants (4.50 g) equals the total mass of products and the excess reactant (4.50 g).

Step 1: Determine the limiting reactant

First, we need to calculate the number of moles of each reactant. To do this, we will divide the mass of the reactants by their respective molar masses. Molar masses: NH₃: 14.01 (N) + 3 * 1.01 (H) = 17.03 g/mol O₂: 2 * 16.00 (O) = 32.00 g/mol Number of moles of NH₃: \(= \frac {2.00 g}{17.03 g/mol} = 0.1174 \thinspace moles\) Number of moles of O₂: \(= \frac {2.50 g}{32.00 g/mol} = 0.07813 \thinspace moles\) Now that we have the moles of the reactants, we can determine the limiting reactant. According to the balanced chemical equation, 4 moles of NH₃ react with 5 moles of O₂. Therefore, the required mole ratio is: \(= \frac {4 \thinspace moles \thinspace NH_3}{5 \thinspace moles \thinspace O_2}\) Now, we compare the available moles of reactants to this ratio. Divide the available mole of each reactant by its corresponding mole value in the ratio: NH₃: \(= \frac {0.1174}{4} = 0.02935\) O₂: \(= \frac {0.07813}{5} = 0.01563\) Since O₂ has a smaller ratio, it is the limiting reactant.

Step 2: Calculate the formed products' masses

We will use the number of moles of the limiting reactant (O₂) to find out the number of moles of the products formed (NO and H₂O). According to the balanced equation, 5 moles of O₂ react to form 4 moles of NO and 6 moles of H₂O. Number of moles of NO formed: \(= \frac {4 \thinspace moles \thinspace NO}{5 \thinspace moles \thinspace O_2} \times 0.07813 \thinspace moles \thinspace O_2 = 0.06250 \thinspace moles \thinspace NO\) Number of moles of H₂O formed: \(= \frac {6 \thinspace moles \thinspace H_2O}{5 \thinspace moles \thinspace O_2} \times 0.07813 \thinspace moles \thinspace O_2 = 0.09375 \thinspace moles \thinspace H_2O\) Now, multiply these values by their respective molar masses to find the mass of the formed products: Mass of NO formed: \(= 0.06250 \thinspace moles \thinspace NO \times \frac {30.01 \thinspace g}{1 \thinspace mol} = 1.875 \thinspace g \thinspace NO\) Mass of H₂O formed: \(= 0.09375 \thinspace moles \thinspace H_2O \times \frac {18.02 \thinspace g}{1 \thinspace mol} = 1.689 \thinspace g \thinspace H_2O\)

Step 3: Calculate the mass of the excess reactant that remains

We know that O₂ is the limiting reactant, and thus, it will be completely consumed. We will now calculate how many moles of NH₃ were consumed during this reaction: Number of moles of NH₃ consumed: \(= \frac {4 \thinspace moles \thinspace NH_3}{5 \thinspace moles \thinspace O_2} \times 0.07813 \thinspace moles \thinspace O_2 = 0.06250 \thinspace moles \thinspace NH_3\) Now, we will find the number of moles of NH₃ remaining: Number of moles of NH₃ remaining: \(= 0.1174 \thinspace moles \thinspace NH_3 - 0.06250 \thinspace moles \thinspace NH_3 = 0.05490 \thinspace moles \thinspace NH_3\) Finally, we will calculate the mass of NH₃ remaining by multiplying it with the molar mass of NH₃: Mass of NH₃ remaining: \(= 0.05490 \thinspace moles \thinspace NH_3 \times \frac {17.03 \thinspace g}{1 \thinspace mol} = 0.935 \thinspace g \thinspace NH_3\)

Step 4: Verify the conservation of mass

According to the law of conservation of mass, the total mass of reactants should equal the total mass of products and unreacted excess reactant: Mass of reactants: \(= 2.00 \thinspace g \thinspace NH_3 + 2.50 \thinspace g \thinspace O_2 = 4.50 \thinspace g\) Add the masses of formed products NO and H₂O, and unreacted NH₃: Mass of products and remaining excess reactant: \(= 1.875 \thinspace g \thinspace NO + 1.689 \thinspace g \thinspace H_2O + 0.935 \thinspace g \thinspace NH_3 = 4.50 \thinspace g\) Since the total mass of reactants equals the total mass of products and the excess reactant, our calculations are consistent with the law of conservation of mass.

Key Concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that runs out first, stopping the reaction from continuing. To find the limiting reactant, we calculate the moles of each reactant based on their given mass and compare their proportion to the stoichiometric ratio from the balanced equation. For instance, in the conversion of ammonia (\(\text{NH}_3\)) to nitric oxide (\(\text{NO}\)), we have the reaction:
4 \(\text{NH}_3 \) + 5 \( \text{O}_2 \) → 4 \( \text{NO} \) + 6 \( \text{H}_2 \text{O} \).
We first determine the moles:

• For \( \text{NH}_3 \): Mass = 2.00 g, Molar mass = 17.03 g/mol, Moles = 0.1174 moles
• For \( \text{O}_2 \): Mass = 2.50 g, Molar mass = 32.00 g/mol, Moles = 0.07813 moles

The stoichiometric ratio demands more oxygen per mole than ammonia. Based on available moles, \( \text{O}_2 \) has the smaller ratio (0.01563), making it the limiting reactant. As the limiting reactant runs out, it determines the amount of product formed, restricting the reaction's completion.

Conservation of Mass

The Law of Conservation of Mass states that mass cannot be created or destroyed in a chemical reaction. This principle is fundamental in stoichiometry, ensuring that the mass of reactants equals the mass of products and leftover substances. After the reaction between ammonia and oxygen, we calculate mass to verify this law. The masses before and after the reaction must be the same. Let's look at the experiment:

• Total mass of reactants = Mass of \( \text{NH}_3 \) + Mass of \(\text{O}_2\) = 2.00 g + 2.50 g = 4.50 g.
• Total mass of products and excess reactant: Mass of \( \text{NO} \) + Mass of \(\text{H}_2\text{O}\) + excess \(\text{NH}_3\) = 1.875 g + 1.689 g + 0.935 g = 4.50 g.

The sum of the product masses and unreacted reactants matches the total reactant mass, confirming mass conservation. No mass is gained or lost, adhering to the conservation law in every chemical process.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions using the balanced chemical equation. It is based on the mole concept and molar ratios. This concept helps predict the quantities of substances consumed and produced. It involves several steps:
First, determine moles of all reactants using their masses and molar masses. Next, use the balanced equation to find mole ratios. During ammonia's reaction with oxygen to form nitric oxide and water, we use stoichiometry to compute:

• From 0.07813 moles of \(\text{O}_2\), how many moles of \(\text{NO}\) and \(\text{H}_2\text{O}\) are formed?
• Use ratios from the equation: 5 moles \(\text{O}_2\) forms 4 moles \(\text{NO}\) and 6 moles \(\text{H}_2\text{O}.\)
• Calculate mass of products: \(\text{NO}\) = 1.875 g, \(\text{H}_2\text{O}\) = 1.689 g,

Stoichiometry ensures all calculations are in balance, making it a vital tool in predicting precise amounts in chemical reactions.