Question

For cach of the following balanced chemical cquations, calculate how many moles and how many grams of each product would be produced by the complete conversion of 0.50 mole of the reactant indicated in boldface. State clearly the mole ratio used for each conversion. a. \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) b. \(\mathrm{CH}_{4}(g)+4 \mathrm{~S}(s) \rightarrow \mathrm{CS}_{2}(l)+2 \mathrm{H}_{2} \mathrm{~S}(g)\) c. \(\mathbf{P C l}_{3}(l)+3 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+3 \mathrm{HCl}(a q)\) d. \(\mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{NaHCO}_{3}(s)\)

Short answer

Short answer: a. \(0.50\) moles and \(26.75\) grams of NH4Cl were produced. b. \(0.50\) moles and \(38.08\) grams of CS2, and \(1.0\) mole and \(34.09\) grams of H2S were produced. c. \(0.50\) moles and \(40.99\) grams of H3PO3, and \(1.5\) moles and \(54.69\) grams of HCl were produced. d. \(0.50\) moles and \(42.01\) grams of NaHCO3 were produced.

Step-by-step solution

Identify moles of reactants and mole ratio of conversion

We have 0.50 mole of NH3, and the mole ratio between NH3 and NH4Cl is 1:1 (from the balanced equation).

Calculate moles of NH4Cl produced

Using the 1:1 mole ratio, we know that the same number of moles of NH4Cl will be produced as the moles of NH3 we have, which is 0.50 moles.

Calculate mass of NH4Cl produced

To find the mass of NH4Cl produced, we need to multiply its moles by its molar mass. The molar mass of NH4Cl = 14.01 (N) + 1.01 * 4 (H) + 35.45 (Cl) = 53.49 g/mol. Therefore, the mass of NH4Cl produced = 0.50 moles * 53.49 g/mol = 26.75 g. ###Problem b: CH4(g) + 4 S(s) -> CS2(l) + 2 H2S(g)

Identify moles of reactants and mole ratios of conversion

We have 0.50 mole of CH4, and the mole ratios between CH4 and CS2 and between CH4 and H2S are 1:1 and 1:2, respectively (from the balanced equation).

Calculate moles of CS2 and H2S produced

Using the 1:1 mole ratio, moles of CS2 produced = 0.50 moles. Using the 1:2 mole ratio, moles of H2S produced = 0.50 moles * 2 = 1.0 mole.

Calculate masses of CS2 and H2S produced

Molar mass of CS2 = 12.01 (C) + 32.07 * 2 (S) = 76.15 g/mol. Mass of CS2 produced = 0.50 moles * 76.15 g/mol = 38.08 g. Molar mass of H2S = 1.01 * 2 (H) + 32.07 (S) = 34.09 g/mol. Mass of H2S produced = 1.0 mole * 34.09 g/mol = 34.09 g. ###Problem c: PCl3(l) + 3 H2O(l) -> H3PO3(aq) + 3 HCl(aq)

Identify moles of reactants and mole ratios of conversion

We have 0.50 mole of PCl3, and the mole ratios between PCl3 and H3PO3 and between PCl3 and HCl are 1:1 and 1:3, respectively (from the balanced equation).

Calculate moles of H3PO3 and HCl produced

Using the 1:1 mole ratio, moles of H3PO3 produced = 0.50 moles. Using the 1:3 mole ratio, moles of HCl produced = 0.50 moles * 3 = 1.5 moles.

Calculate masses of H3PO3 and HCl produced

Molar mass of H3PO3 = 1.01 * 3 (H) + 30.97 (P) + 16.00 * 3 (O) = 81.98 g/mol. Mass of H3PO3 produced = 0.50 moles * 81.98 g/mol = 40.99 g. Molar mass of HCl = 1.01 (H) + 35.45 (Cl) = 36.46 g/mol. Mass of HCl produced = 1.5 moles * 36.46 g/mol = 54.69 g. ###Problem d: NaOH(s) + CO2(g) -> NaHCO3(s)

Identify moles of reactants and mole ratio of conversion

We have 0.50 mole of NaOH, and the mole ratio between NaOH and NaHCO3 is 1:1 (from the balanced equation).

Calculate moles of NaHCO3 produced

Using the 1:1 mole ratio, we know that the same number of moles of NaHCO3 will be produced as the moles of NaOH we have, which is 0.50 moles.

Calculate mass of NaHCO3 produced

Molar mass of NaHCO3 = 22.99 (Na) + 1.01 (H) + 12.01 (C) + 16.00 * 3 (O) = 84.01 g/mol. Mass of NaHCO3 produced = 0.50 moles * 84.01 g/mol = 42.01 g.

Key concepts

Chemical Reactions

Chemical reactions are processes where starting substances, called reactants, undergo transformations to become new substances known as products. These reactions are typically depicted by balanced chemical equations. For instance, in the chemical equation \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\), ammonia \((\mathrm{NH}_{3})\) and hydrochloric acid \((\mathrm{HCl})\) react to form ammonium chloride \((\mathrm{NH}_{4}\mathrm{Cl})\).

Understanding the nature of these reactions involves knowing which bonds are broken and formed during the reaction. This helps in the visualization of the process and predicts the outcome in terms of products. In this context:

• The balanced equation ensures that the number of each type of atom is the same on both sides of the equation, following the Law of Conservation of Mass.
• Each equation represents a specific stoichiometric relationship between reactants and products, which is essential for quantitative analysis.

By examining these relationships, one can calculate how much product will form from a given amount of reactant.

Mole Concept

The mole concept is a fundamental aspect of chemistry that relates the mass of substances to the quantity of atoms, molecules, or formula units they contain. A mole is a unit that represents \(6.022 \times 10^{23}\) of these entities, which is known as Avogadro's number.

For example, if you're working with \(0.50\) mole of \(\mathbf{PCl}_{3}\), it represents \(0.50 \times 6.022 \times 10^{23}\) molecules of \(\mathrm{PCl}_{3}\). The mole provides a bridge between the microscopic scale of atoms and the macroscopic scale of grams that we measure.

• It allows chemists to convert between the mass of a substance and the number of particles it contains.
• Using moles makes calculations involving chemical reactions manageable and comprehensible.

By applying the mole ratios derived from balanced chemical equations, you can determine the respective amounts of reactants and products, such as in conversion problems where given moles of reactants can predict moles of products formed.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). It plays a crucial role in converting between the mass of a substance and the amount in moles. For instance, to find the molar mass of \(\mathrm{NH}_{4} \mathrm{Cl}\), add the atomic masses of nitrogen \((\mathrm{N})\), hydrogen \((\mathrm{H})\), and chlorine \((\mathrm{Cl})\):

\[\text{Molar Mass of } \mathrm{NH}_{4} \mathrm{Cl} = 14.01 + 4 \times 1.01 + 35.45 = 53.49 \text{ g/mol}\]

This calculation facilitates understanding of how much a substance weighs in terms of grams per mole.

• Molar mass empowers conversions between mass and moles, allowing for precise stoichiometric calculations in chemical equations.
• It is particularly essential when predicting the mass of products formed or reactants required in a chemical reaction, as shown in the example: \(0.50\) moles of \(\mathrm{NH}_{4} \mathrm{Cl}\) produced is equivalent to \(0.50 \times 53.49\), resulting in \(26.75\) grams.

By mastering molar mass calculations, students can unlock insights into the practical aspects of both theoretical and applied chemistry.