Question

The reaction of \(750 .\) g each of \(\mathrm{NH}_{3}\) and \(\mathrm{O}_{2}\) was found to produce \(562 \mathrm{g}\) of \(\mathrm{NO}\) (see pages \(153-155\) ). $$ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ (a) What mass of water is produced by this reaction? (b) What quantity of \(\mathrm{O}_{2}\) is required to consume \(750 .\) g of \(\mathrm{NH}_{3} ?\)

Short answer

(a) 506.94 g of water is produced. (b) 1762.56 g of O_2 is needed.

Step-by-step solution

Determine moles of reactants

First, calculate the moles of NH_3 and O_2 using their molar masses. The molar mass of NH_3 is approximately 17.03 g/mol and O_2 is 32.00 g/mol. \[ \text{Moles of } NH_3 = \frac{750}{17.03} \approx 44.04 \text{ moles}\] \[ \text{Moles of } O_2 = \frac{750}{32.00} \approx 23.44 \text{ moles} \]

Identify the limiting reactant

Using stoichiometry, determine which reactant runs out first. From the balanced equation, 4 moles of NH_3 react with 5 moles of O_2. Calculate the required moles of O_2 for 44.04 moles of NH_3:\[ \text{Required O}_2 = \frac{44.04 \times 5}{4} = 55.05 \text{ moles} \] Since we only have 23.44 moles of O_2, O_2 is the limiting reactant.

Calculate moles of products based on limiting reactant

Using the stoichiometry of the balanced chemical equation, calculate the number of moles of NO and H_2O produced by the reaction. Since O_2 is the limiting reactant, we use its amount:\[ \text{Moles of NO} = \frac{23.44 \times 4}{5} = 18.75 \text{ moles} \] \[ \text{Moles of H}_2\text{O} = \frac{23.44 \times 6}{5} = 28.128 \text{ moles} \]

Convert moles of NO to mass

Convert the moles of NO to mass using its molar mass of 30.01 g/mol:\[ \text{Mass of NO} = 18.75 \times 30.01 = 562.6875 \text{ g} \] This verifies the mass of NO produced given in the problem.

Convert moles of H2O to mass

Convert the moles of water to mass using its molar mass of 18.015 g/mol:\[ \text{Mass of H}_2\text{O} = 28.128 \times 18.015 = 506.94 \text{ g} \] This is the mass of water produced in the reaction.

Calculate required O2 for 750 g NH3

Given that NH_3 is not the limiting reactant, calculate how much O_2 is needed to react completely with 44.04 moles of NH_3:\[ \text{Required O}_2 = \frac{44.04 \times 5}{4} \times 32.00 = 1762.56 \text{ g} \] This is the mass of O_2 needed to completely react with 750 g of NH_3.

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that gets used up first and limits the extent of the reaction. This reactant determines the amount of products formed. Understanding which reactant is limiting is crucial because it helps predict the quantities of products that can be produced in a reaction.
For this exercise, when 750 grams of both H_3 m{NH}_3 m{ and O_2 m{O}_2 m{ were used, we had to determine which one would run out first. Using stoichiometry, we calculated that we needed 55.05 moles of O_2 m{O}_2 m{ for the reaction with 44.04 moles of H_3 m{NH}_3 m{. However, we only had 23.44 moles of O_2 m{O}_2 m{ available, making it the limiting reactant.
Thus, O_2 m{O}_2 m{ dictates how far this reaction could proceed, impacting the quantity of NO m{NO m{ and H_2 m{H}_2 m{O m{ that could be produced.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, often through the breaking and forming of chemical bonds. These reactions are represented by chemical equations like the one given in the exercise: 4 H_3 m{NH}_3 m{(g) + 5 O_2 m{O}_2 m{(g) → 4 O m{NO m{(g) + 6 H_2 m{H}_2 m{O m{(g). In this balanced equation, ammonia ( H_3 m{NH}_3 m{) reacts with oxygen ( O_2 m{O}_2 m{) to produce nitrogen monoxide ( O m{NO m{) and water ( H_2 m{H}_2 m{O m{).
Reactants undergo a chemical change, leading to the formation of products. The reaction might release or absorb energy, with products potentially having different physical and chemical properties from the reactants. Understanding these reactions is key to predicting how substances interact and transform in chemical processes.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole. It acts as a bridge between the macroscopic world of grams and the microscopic world of atoms and molecules. To compute the molar mass, you sum up the atomic masses of the elements in the molecule, considering the number of each type of atom present.

• For H_3 m{NH}_3 m{, with a molar mass of 17.03 g/mol.
• For O_2 m{O}_2 m{, with a molar mass of 32.00 g/mol.

In this exercise, knowing the molar masses of H_3 m{NH}_3 m{ and O_2 m{O}_2 m{ allowed us to convert from grams to moles. It facilitated the analysis of how much reactant was available and subsequently how much product could be formed. Ultimately, it helps to detail how much of each reactant is needed to balance the equation and complete the reaction.

Balanced Chemical Equation

A balanced chemical equation reflects the principle of the conservation of mass, stating that matter is neither created nor destroyed in a chemical reaction. This means the number of atoms for each element must be equal on both sides of the equation.
For instance, in the balanced equation of the exercise: 4 H_3 m{NH}_3 m{(g) + 5 O_2 m{O}_2 m{(g) → 4 O m{NO m{(g) + 6 H_2 m{H}_2 m{O m{(g), there are 4 nitrogen atoms, 12 hydrogen atoms, and 10 oxygen atoms on both sides. This ensures that all elements are accounted for and conserved through the reaction.

• The coefficients indicate the proportions needed for the reactants to transform into products.
• This is vital for calculating theoretical yields and for determining the limiting reactant.

Balancing equations is foundational for accurately predicting the outcomes of reactions and is an essential step in solving stoichiometry problems.