Question

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly into the wound. If \(25.0 \mathrm{~g} \mathrm{Ag}_{2} \mathrm{O}\) is reacted with \(50.0 \mathrm{~g} \mathrm{C}_{10} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{SO}_{2}\), what mass of silver sulfadiazine, \(\mathrm{AgC}_{10} \mathrm{H}_{9} \mathrm{~N}_{4} \mathrm{SO}_{2}\), can be produced, assuming \(100 \%\) yield? \(\mathrm{Ag}_{2} \mathrm{O}(s)+2 \mathrm{C}_{10} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{SO}_{2}(s) \longrightarrow 2 \mathrm{AgC}_{10} \mathrm{H}_{9} \mathrm{~N}_{4} \mathrm{SO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

When 25.0g of Ag2O reacts with 50.0g of C10H10N4SO2, assuming a 100% yield, \(71.33 \ g\) of silver sulfadiazine (AgC10H9N4SO2) can be produced.

Step-by-step solution

Convert Masses to Moles

To convert the masses of the reactants to moles, we first need to determine the molar mass of each reactant: Molar mass of Ag2O = \(2 * Ag + O = 2 * 107.87 + 16.00 = 231.74 \ g/mol\) Molar mass of C10H10N4SO2 = \(10 * C + 10 * H + 4 * N + S + 2 * O = 10 * 12.01 + 10 * 1.01 + 4 * 14.01 + 32.07 + 2 * 16.00 = 250.30 \ g/mol\) Now, we can convert the masses to moles using the molar mass for each reactant: Moles of Ag2O = \(25.0g \div 231.74g/mol = 0.1078 \ mol\) Moles of C10H10N4SO2 = \(50.0g \div 250.30g/mol = 0.1997 \ mol\)

Determine the Limiting Reactant

Using the balanced chemical equation, we can determine the mole ratio between the reactants: Mole ratio (Ag2O:C10H10N4SO2) = 1:2 By dividing the moles of each reactant by their respective mole ratio, we can determine which reactant is the limiting reactant: For Ag2O: \(0.1078 \ mol ÷ 1 = 0.1078\) For C10H10N4SO2: \(0.1997 \ mol ÷ 2 = 0.09985\) Since 0.09985 is smaller than 0.1078, C10H10N4SO2 is the limiting reactant.

Calculate Moles of Silver Sulfadiazine Produced

Using the stoichiometry from the balanced chemical equation, we can calculate the moles of silver sulfadiazine (AgC10H9N4SO2) produced: Mole ratio (AgC10H9N4SO2:C10H10N4SO2) = 2:2 Moles of AgC10H9N4SO2 produced = \(0.1997 \ mol \times \frac{2 \ mol \ AgC10H9N4SO2}{2 \ mol \ C10H10N4SO2} = 0.1997\ mol\)

Convert Moles of Silver Sulfadiazine to Mass

Now, we need to determine the molar mass of AgC10H9N4SO2: Molar mass of AgC10H9N4SO2 = \(Ag + 10 * C + 9 * H + 4 * N + S + 2 * O = 107.87 + 10 * 12.01 + 9 * 1.01 + 4 * 14.01 + 32.07 + 2 * 16.00 = 357.14 \ g/mol\) Using the molar mass, we can convert the moles of silver sulfadiazine produced into mass: Mass of AgC10H9N4SO2 = \(0.1997 \ mol \times 357.14 \ g/mol = 71.33 \ g\) Therefore, when 25.0g of Ag2O reacts with 50.0g of C10H10N4SO2, assuming a 100% yield, 71.33g of silver sulfadiazine (AgC10H9N4SO2) can be produced.

Key concepts

Limiting Reactant

The limiting reactant in a chemical reaction is the substance that is completely consumed first. It determines the maximum amount of product that can be formed from the given quantities of reactants. In the given exercise, you must identify which reactant limits the production of silver sulfadiazine by comparing the molar ratios.Using the balanced chemical equation: \[\text{Ag}_2\text{O} + 2 \text{C}_{10}\text{H}_{10}\text{N}_{4}\text{SO}_{2} \rightarrow 2 \text{AgC}_{10}\text{H}_{9}\text{N}_{4}\text{SO}_{2} + \text{H}_2\text{O} \]we see the mole ratio between \(\text{Ag}_2\text{O}\) and \(\text{C}_{10}\text{H}_{10}\text{N}_{4}\text{SO}_{2}\) is 1:2. By calculating the moles of each reactant, you can identify which is limiting. Since \(0.09985\), derived from \(\text{C}_{10}\text{H}_{10}\text{N}_{4}\text{SO}_{2}\) is less than \(0.1078\), derived from \(\text{Ag}_2\text{O}\), \(\text{C}_{10}\text{H}_{10}\text{N}_{4}\text{SO}_{2}\) is the limiting reactant.Understanding the concept of limiting reactants helps prevent waste and optimizes chemical reactions, essential for industrial processes.

Molar Mass

Molar mass is a critical quantity in chemistry used to convert between mass and moles. It is the mass of one mole of a given substance and is expressed in units of grams per mole (g/mol).To find the molar mass, sum the atomic masses of all the atoms in a molecule. For example:- For \(\text{Ag}_2\text{O}\): Molar mass = \(2 \times \text{Ag} + \text{O} = 2 \times 107.87 + 16.00 = 231.74 \text{ g/mol}\)- For \(\text{C}_{10}\text{H}_{10}\text{N}_{4}\text{SO}_{2}\): Molar mass = \(10 \times \text{C} + 10 \times \text{H} + 4 \times \text{N} + \text{S} + 2 \times \text{O} = 250.30 \text{ g/mol}\)Understanding how to compute molar masses allows you to perform conversions from grams to moles, facilitating stoichiometric calculations in reactions, like determining the amount of product formed.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions, displaying reactants converting to products. They must be balanced to reflect the conservation of mass, meaning the number of atoms of each element should be equal on both sides.For the equation presented:\[\text{Ag}_2\text{O} + 2 \text{C}_{10}\text{H}_{10}\text{N}_{4}\text{SO}_{2} \rightarrow 2 \text{AgC}_{10}\text{H}_{9}\text{N}_{4}\text{SO}_{2} + \text{H}_2\text{O} \]Each side contains the same number of each type of atom. Balance ensures reactions comply with the law of conservation of mass.Moreover, chemical equations enable the determination of reactant ratios and calculations of theoretical yields through stoichiometry, providing insights into reaction efficiency and required reactant quantities.

Yield Calculation

Yield calculation is a practical measure in chemistry to determine how much product is formed in a reaction. The yield is often expressed as a percentage, comparing the actual amount of product obtained to the theoretical maximum amount predicted by stoichiometry.In the given example, the task is to calculate the mass of silver sulfadiazine assuming 100% yield. This means all the limiting reactants get converted into the desired product without losses. Using the moles of produced \(\text{AgC}_{10}\text{H}_{9}\text{N}_{4}\text{SO}_{2}\), the mass is calculated as:\[0.1997 \text{ mol} \times 357.14 \text{ g/mol} = 71.33 \text{ g} \]In manufacturing and laboratory settings, yield is crucial to evaluating efficiency. A 100% yield is ideal but often not achievable due to practical constraints. Real-world yields help in assessing reaction conditions and in scaling processes for industrial applications.