Question

What mass of solid \(\mathrm{AgBr}\) is produced when \(100.0 \mathrm{~mL}\) of \(0.150 \mathrm{MAgNO}_{3}\) is added to \(20.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{NaBr} ?\)

Short answer

When \(100.0\) mL of \(0.150\) M \(\mathrm{AgNO}_{3}\) is added to \(20.0\) mL of \(1.00\) M \(\mathrm{NaBr}\), \(2.82\) g of solid \(\mathrm{AgBr}\) is produced.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction between silver nitrate and sodium bromide is: \(\mathrm{AgNO}_{3} (aq) + \mathrm{NaBr} (aq) \rightarrow \mathrm{AgBr} (s) + \mathrm{NaNO}_{3} (aq)\)

Calculate the moles of the reactants

Use the molarity and volume of each solution to calculate the moles of each reactant: Moles of \(\mathrm{AgNO}_{3}\): moles = Molarity × Volume moles = \(0.150 \mathrm{M} \times 100.0 \mathrm{mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 0.0150 \, mol\) Moles of \(\mathrm{NaBr}\): moles = Molarity × Volume moles = \(1.00 \mathrm{M} \times 20.0 \mathrm{mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 0.0200 \, mol\)

Determine the limiting reactant

To find the limiting reactant, compare the mole ratio of the reactants to the stoichiometry of the balanced equation (in this case, it's a 1:1 ratio): Mole ratio = \(\frac{moles \, of \, \mathrm{AgNO}_{3}}{moles \, of \, \mathrm{NaBr}} = \frac{0.0150 \, mol}{0.0200 \, mol} = 0.750\) Since the mole ratio is less than 1, \(\mathrm{AgNO}_{3}\) is the limiting reactant. This means that all of the \(\mathrm{AgNO}_{3}\) will be used up in the reaction.

Calculate the mass of solid \(\mathrm{AgBr}\) produced

Using the moles of the limiting reactant and the stoichiometry of the balanced equation, calculate the number of moles of \(\mathrm{AgBr}\) produced: moles of \(\mathrm{AgBr}\) = moles of \(\mathrm{AgNO}_{3}\) = \(0.0150 \, mol\) Now, we'll convert the moles of \(\mathrm{AgBr}\) into mass using its molar mass (\(107.87 \mathrm{g/mol} + 79.9 \mathrm{g/mol} = 187.77 \mathrm{g/mol}\)): mass of \(\mathrm{AgBr}\) = moles of $\mathrm{AgBr} \times \mathrm{molar \, mass \, of} \, \mathrm{AgBr}\) mass = $0.0150 \, mol \times 187.77 \mathrm{g/mol} = 2.82 \, g\) So, when \(100.0\) mL of \(0.150\) M \(\mathrm{AgNO}_{3}\) is added to \(20.0\) mL of \(1.00\) M \(\mathrm{NaBr}\), \(2.82\) g of solid \(\mathrm{AgBr}\) is produced.

Key concepts

Limiting Reactant

Understanding the limiting reactant is essential in solving many stoichiometry problems. The limiting reactant is the substance in a chemical reaction that is completely consumed first, thus determining the maximum amount of product that can be formed. In a typical reaction, when the reactants are not present in the exact stoichiometric ratio consistent with the balanced equation, one will limit the extent of the reaction.

The amount of product formed is directly proportional to the amount of the limiting reactant. To identify the limiting reactant, you must first calculate the moles of each reactant. Then, compare these moles with the mole ratio from the balanced chemical equation. The reactant which has the lower mole ratio in comparison to the balanced equation is the limiting reactant, as seen in the provided exercise where \(\mathrm{AgNO_{3}}\) is identified as the limiting reactant.

Molar Mass Calculation

Molar mass is a critical factor in converting between moles and grams and vice versa. The molar mass is the weight of one mole of a substance and is measured in grams per mole (g/mol). It can be calculated by summing the atomic masses of all the atoms in the molecular formula of the substance.

For each element, you multiply the atomic mass by the number of atoms of that element in the formula, and then add the results for all elements present. When determining the mass of the product in a chemical reaction, you’ll need the molar mass to convert from moles to grams. An example is seen in the original exercise where the molar mass of \(\mathrm{AgBr}\) is obtained by adding the atomic masses of silver (Ag) and bromine (Br) to find 187.77 g/mol.

Mole-to-Mass Conversion

Converting moles to mass is a frequent procedure in stoichiometry which involves multiplying the molar mass of a substance by the number of moles present. This conversion is based on the definition of a mole, which is 6.022 x 10^{23} particles of the substance (Avogadro’s number), synonymous to the molar mass in grams.

In practice, after identifying the limiting reactant and using stoichiometry to determine the moles of product produced, you then use the molar mass of the product to find the mass in grams. This process, exemplified in the exercise solution, shows how understanding the relationship between moles and mass is crucial for predicting the mass of products formed in a chemical reaction.