Question

For each of the following balanced equations, indicate how many moles of the product could be produced by complete reaction of \(1.00 \mathrm{g}\) of the reactant indicated in boldface. Indicate clearly the mole ratio used for the conversion. a. \(\mathbf{N H}_{3}(g)+\mathrm{HCl}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) b. \(\mathbf{C a O}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CaCO}_{3}(s)\) c. \(\mathbf{4} \mathbf{N} \mathbf{a}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Na}_{2} \mathrm{O}(s)\) d. \(\mathbf{2} \mathbf{P}(s)+3 \mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{PCl}_{3}(l)\)

Short answer

a. 0.0588 moles of NH4Cl are produced (mole ratio 1:1). b. 0.0179 moles of CaCO3 are produced (mole ratio 1:1). c. 0.0217 moles of Na2O are produced (mole ratio 2:1). d. 0.0323 moles of PCl3 are produced (mole ratio 1:1).

Step-by-step solution

a. NH3(g) + HCl(g) → NH4Cl(s)

First, we need to find the moles of NH3. The molar mass of NH3 = 14 (N) + 3(1) (H) = 17g/mol. So, moles of NH3 = mass / molar mass = 1.00g / 17g/mol = 0.0588 moles Now we can use the mole ratio between NH3 and NH4Cl, which is 1:1. Thus, moles of NH4Cl formed = moles of NH3 = 0.0588 moles The mole ratio used is 1:1.

b. CaO(s) + CO2(g) → CaCO3(s)

First, we need to find the moles of CaO. The molar mass of CaO = 40 (Ca) + 16 (O) = 56g/mol. So, moles of CaO = mass / molar mass = 1.00g / 56g/mol = 0.0179 moles Now we can use the mole ratio between CaO and CaCO3, which is 1:1. Thus, moles of CaCO3 formed = moles of CaO = 0.0179 moles The mole ratio used is 1:1.

c. 4Na(s) + O2(g) → 2Na2O(s)

First, we need to find the moles of Na. The molar mass of Na = 23g/mol. So, moles of Na = mass / molar mass = 1.00g / 23g/mol = 0.0435 moles Now we can use the mole ratio between Na and Na2O, which is 4:2 or 2:1. Thus, moles of Na2O formed = (1/2) × moles of Na = (1/2) × 0.0435 = 0.0217 moles The mole ratio used is 2:1.

d. 2P(s) + 3Cl2(g) → 2PCl3(l)

First, we need to find the moles of P. The molar mass of P = 31g/mol. So, moles of P = mass / molar mass = 1.00g / 31g/mol = 0.0323 moles Now we can use the mole ratio between P and PCl3, which is 2:2 or 1:1. Thus, moles of PCl3 formed = moles of P = 0.0323 moles The mole ratio used is 1:1.

Key concepts

Mole Ratio

The mole ratio, a fundamental aspect of stoichiometry, is the ratio of moles of one substance to the moles of another substance in a balanced chemical equation. It is derived directly from the coefficients of the substances involved in the reaction. Understanding mole ratio is crucial for converting moles from one substance to another in chemical reactions.

For instance, in the reaction between ammonia (NH3) and hydrochloric acid (HCl) to form ammonium chloride (NH4Cl), the equation shows a 1:1 ratio. This means for every 1 mole of NH3 that reacts, 1 mole of NH4Cl is produced. Similarly, in the reaction of calcium oxide (CaO) with carbon dioxide (CO2) to form calcium carbonate (CaCO3), the mole ratio is also 1:1. Recognizing these ratios allows us to predict the amount of product formed from a certain amount of reactant.

In more complex reactions, such as the formation of sodium oxide (Na2O) from sodium (Na) and oxygen (O2), we observe a different mole ratio. Here, 4 moles of sodium react with 1 mole of oxygen to produce 2 moles of sodium oxide, resulting in a 4:2 or simplified 2:1 ratio for Na to Na2O. In every one of these examples, the mole ratio serves as the conversion factor that underpins the stoichiometric calculations.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Each element on the periodic table has a unique molar mass; for example, nitrogen (N) has a molar mass of approximately 14 g/mol, while hydrogen (H) is about 1 g/mol. The molar mass of a compound, like ammonia (NH3), is calculated by adding up the molar masses of its constituent elements.

The calculation of molar mass is essential for converting between the mass of a substance and the amount in moles. When solving stoichiometry problems, determining the molar mass allows us to start the conversion process. For example, to find the number of moles of ammonia when provided its mass, we divide the mass by its molar mass, thus 1.00 g divided by 17 g/mol yields 0.0588 moles of NH3.

Practical Applications

In a real-world scenario, knowing molar mass can help in preparing solutions with precise concentrations or determining how much of a reactant is needed to fully react with a given quantity of another reactant. It is a bridge between the microscopic world of atoms and molecules and the macroscopic world that we can measure and observe.

Balanced Chemical Equations

Balanced chemical equations are the cornerstone of chemical stoichiometry. They express the principle of conservation of mass, indicating that the number of atoms of each element involved in a reaction remains constant, simply rearranged to form new substances.

A chemical equation is balanced when the number of atoms of each element on the reactant side equals the number of atoms on the product side. For example, the equation for the reaction of sodium (Na) with oxygen (O2) to form sodium oxide (Na2O) is balanced as 4 Na + O2 → 2 Na2O; it shows that four sodium atoms react with one diatomic oxygen molecule to produce two units of sodium oxide, each composed of two sodium atoms and one oxygen atom.

Why Balancing Equations Matters

Without a balanced equation, calculations involving mole ratios would not accurately reflect the stoichiometry of the reaction. This balance ensures that the mole ratios deduced from it can reliably be used to calculate the amounts of reactants needed and products formed, making it essential for both laboratory work and industry processes. Accurate stoichiometric calculations depend on the foundation of correctly balanced chemical equations.