Question

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

Short answer

a. Using a 1:1 mole ratio, 0.50 mol of \(NH_3\) produces 0.50 mol of \(NH_4Cl\), resulting in 26.745 g of \(NH_4Cl\). b. With 0.50 mol of \(S\), 0.125 mol of \(CS_2\) (9.518 g) and 0.25 mol of \(H_2S\) (8.52 g) are produced, with mole ratios of 4:1 and 4:2, respectively. c. From 0.50 mol of \(PCI_3\), 0.50 mol of \(H_3PO_3\) (40.995 g) and 1.50 mol of \(HCl\) (54.69 g) are produced, with a 1:1 and 1:3 mole ratio, respectively. d. Using a 1:1 mole ratio, 0.50 mol of \(NaOH\) produces 0.50 mol of \(NaHCO_3\), resulting in 42.005 g of \(NaHCO_3\).

Step-by-step solution

Identify the mole ratio

In this equation, the mole ratio is 1:1 for all reactants and products.

Calculate the moles of product

Since 0.50 mol of \(NH_3\) is reacting, and the mole ratio is 1:1, we will get 0.50 mol of \(NH_4Cl\).

Calculate the grams of product

To convert moles to grams, we'll multiply the moles by the molar mass of \(NH_4Cl\), which is 53.49 g/mol. 0.50 mol × 53.49 g/mol = 26.745 g So, 26.745 g of \(NH_4Cl\) will be produced. b. \(CH_{4}(g) + 4S(s) \rightarrow CS_{2}(l) + 2H_{2}S(g)\)

Identify the mole ratio

In this equation, the mole ratio is 1:4:1:2 for \(CH_4:S:CS_2:H_2S\).

Calculate the moles of product

Since 0.50 mol of \(S\) is reacting, and the mole ratio with \(CS_2\) is 4:1, we will get 0.125 mol of \(CS_2\) and with \(H_2S\) being 4:2, we will get 0.25 mol of \(H_2S\).

Calculate the grams of product

To convert moles to grams, we'll multiply the moles by their respective molar masses. The molar mass of \(CS_2\) is 76.14 g/mol, and the molar mass of \(H_2S\) is 34.08 g/mol. 0.125 mol × 76.14 g/mol = 9.518 g of \(CS_2\) 0.25 mol × 34.08 g/mol = 8.52 g of \(H_2S\) So, 9.518 g of \(CS_2\) and 8.52 g of \(H_2S\) will be produced. c. \(PCI_{3} + 3H_{2}O(l) \rightarrow H_{3}PO_{3}(aq) + 3HCl(aq)\)

Identify the mole ratio

In this equation, the mole ratio is 1:3:1:3 for \(PCI_3:H_2O:H_3PO_3:HCl\).

Calculate the moles of product

Since 0.50 mol of \(PCI_3\) is reacting, and the mole ratio is 1:1, we will get 0.50 mol of \(H_3PO_3\) and 1.50 mol of \(HCl\).

Calculate the grams of product

To convert moles to grams, we'll multiply the moles by their respective molar masses. The molar mass of \(H_3PO_3\) is 81.99 g/mol, and the molar mass of \(HCl\) is 36.46 g/mol. 0.50 mol × 81.99 g/mol = 40.995 g of \(H_3PO_3\) 1.50 mol × 36.46 g/mol = 54.69 g of \(HCl\) So, 40.995 g of \(H_3PO_3\) and 54.69 g of \(HCl\) will be produced. d. \(NaOH(s) + CO_{2}(g) \rightarrow NaHCO_{3}(s)\)

Identify the mole ratio

In this equation, the mole ratio is 1:1:1 for \(NaOH:CO_2:NaHCO_3\).

Calculate the moles of product

Since 0.50 mol of \(NaOH\) is reacting, and the mole ratio is 1:1, we will get 0.50 mol of \(NaHCO_3\).

Calculate the grams of product

To convert moles to grams, we'll multiply the moles by the molar mass of \(NaHCO_3\), which is 84.01 g/mol. 0.50 mol × 84.01 g/mol = 42.005 g So, 42.005 g of \(NaHCO_3\) will be produced.

Key concepts

Mole Ratio

A mole ratio is an essential concept in stoichiometry that helps chemists understand the relationship between different substances in a balanced chemical equation. It's essentially a proportional comparison between the amounts of reactants and products involved in chemical reactions.

To determine the mole ratio, you examine the coefficients in the balanced chemical equation. These coefficients represent the moles of each substance reacting or being formed. For instance, in the equation \(NH_3 + HCl \rightarrow NH_4Cl\), the mole ratio is 1:1 for both reactants and the product. This means that if one mole of \(NH_3\) reacts, one mole of \(NH_4Cl\) will be produced.

Understanding mole ratios is crucial because it allows for the conversion of moles of a starting substance into moles of another substance. For example, knowing that the mole ratio of \(CH_4\) to \(CS_2\) is 1:1 enables us to predict that 0.50 moles of sulfur will yield 0.125 moles of \(CS_2\), given a 1:4 mole consumption ratio due to the 4:1 original reaction setup.

Molar Mass

Molar mass is a fundamental concept in chemistry, representing the mass of one mole of a given substance. It is expressed in grams per mole (g/mol), and it helps convert between the amount of substance in moles and its mass in grams.

Calculating molar mass involves adding up the atomic masses of all the atoms present in a molecule. For example, the molar mass of \(NH_4Cl\) is calculated by summing the atomic masses of nitrogen (approximately 14.01 g/mol), hydrogen (1.01 g/mol each for four hydrogen atoms), and chlorine (35.45 g/mol), resulting in 53.49 g/mol.

This concept is especially useful in stoichiometry to convert between moles of a substance and its mass, such as finding out that 0.50 moles of \(NH_4Cl\) weigh 26.745 grams. Similar calculations with different molar masses enable us to determine the mass of products in various chemical reactions, such as transforming 0.25 moles of \(H_2S\) into 8.52 grams based on its molar mass of 34.08 g/mol.

Chemical Reactions

Chemical reactions are processes where reactants are transformed into products through the breaking and forming of chemical bonds. Every chemical reaction follows a specific path dictated by the stoichiometry of the equation.

There are various types of chemical reactions including synthesis, decomposition, single replacement, and double replacement. In synthesis reactions, reactants combine to form a single product, such as \(NaOH + CO_2 \rightarrow NaHCO_3\). In this reaction, the 1:1 mole ratio implies that one mole of \(NaOH\) produces one mole of \(NaHCO_3\).

Understanding chemical reactions involves not just recognizing the type but also applying stoichiometry to predict the amounts of products formed from given amounts of reactants. For example, in \(PCI_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl\), knowing that 0.50 moles of \(PCI_3\) produce 0.50 moles of \(H_3PO_3\) and 1.50 moles of \(HCl\) demonstrates how mole ratios and balanced equations guide quantitative predictions in chemical processes.