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The concept of the limiting reactant is essential in stoichiometry. It determines which reactant will be completely used up first in a chemical reaction, thereby limiting the amount of product formed. In our example, we have two reactants: Silver(I) oxide (\(\text{Ag}_2\text{O}\)) and sulfadiazine (\(\text{C}_{10}\text{H}_{10}\text{N}_4\text{SO}_2\)). To identify the limiting reactant, we first need to find the mole ratio between the reactants based on the balanced chemical equation.- The balanced equation shows that 1 mole of Silver(I) oxide reacts with 2 moles of sulfadiazine.- By converting the given masses of the reactants to moles using their respective molar masses, we found that there are 0.1079 moles of Silver(I) oxide and 0.1998 moles of sulfadiazine.- To determine the limiting reactant, these amounts need to be compared against the required stoichiometric ratios. - For Silver(I) oxide: \[ \frac{0.1079 \, \text{mol}}{1} = 0.1079 \] - For sulfadiazine: \[ \frac{0.1998 \, \text{mol}}{2} = 0.0999 \]- The smaller mole ratio corresponds to the limiting reactant. In this reaction, sulfadiazine is the limiting reactant because 0.0999 is less than 0.1079. - This means sulfadiazine will run out first, dictating the quantity of product formed.

Calculating the molar mass is a vital step in stoichiometry, as it allows us to convert between mass and moles, which are essential units in chemical equations. To find molar mass, sum the atomic masses of all atoms in a molecule.

Silver(I) Oxide (\(\text{Ag}_2\text{O}\)

For Silver(I) oxide, we calculate its molar mass by adding the atomic masses of silver and oxygen:- Silver (\(\text{Ag}\): 107.87 g/mol)- Oxygen (\(\text{O}\): 16.00 g/mol)The molar mass of \(\text{Ag}_2\text{O}\) is calculated as:\[ 2 \times 107.87 + 16.00 = 231.74 \, g/mol \]

Sulfadiazole (\(\text{C}_{10}\text{H}_{10}\text{N}_4\text{SO}_2\)

For sulfadiazine, the calculation is more involved, given the numerous atoms:- Carbon (\(\text{C}\): 10 atoms at 12.01 g/mol each)- Hydrogen (\(\text{H}\): 10 atoms at 1.01 g/mol each)- Nitrogen (\(\text{N}\): 4 atoms at 14.01 g/mol each)- Sulfur (\(\text{S}\): 1 atom at 32.07 g/mol)- Oxygen (\(\text{O}\): 2 atoms at 16.00 g/mol each)The molar mass of \(\text{C}_{10}\text{H}_{10}\text{N}_4\text{SO}_2\) is:\[ 10 \times 12.01 + 10 \times 1.01 + 4 \times 14.01 + 32.07 + 2 \times 16.00 = 250.29 \, g/mol \]These calculations allow us to convert the mass of the reactants into moles, aiding in determining the limiting reactant and the yield of the desired product.

Balancing chemical equations is fundamental to stoichiometry, as it ensures that the law of conservation of mass is obeyed. A balanced equation has equal numbers of each type of atom on both sides of the equation.- For our reaction: \[\text{Ag}_2\text{O}(s) + 2\text{C}_{10}\text{H}_{10}\text{N}_4\text{SO}_2(s) \rightarrow 2\text{AgC}_{10}\text{H}_9\text{N}_4\text{SO}_2(s) + \text{H}_2\text{O}(l)\]- The equation shows one mole of Silver(I) oxide reacting with two moles of sulfadiazine, producing two moles of silver sulfadiazine and water.- Each side of the equation must have the same number of each type of atom. Our equation ensures: - 2 Silver (\(\text{Ag}\)) atoms on each side. - 20 Carbon (\(\text{C}\)) atoms: 2 x 10 from sulfadiazine and silver sulfadiazine. - 18 Hydrogen (\(\text{H}\)) atoms total. - 8 Nitrogen (\(\text{N}\)) atoms. - 2 Sulfur (\(\text{S}\)) atoms. - 5 Oxygen (\(\text{O}\)) atoms including ones from water.Appropriate balancing not only follows natural laws but lays the groundwork for accurate stoichiometric calculations, such as the determination of the limiting reactant and computing theoretical yields.