Question

What mass of solid 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

The mass of solid AgBr produced when 100.0 mL of 0.150 M AgNO3 is added to 20.0 mL of 1.00 M NaBr is approximately 2.82 g.

Step-by-step solution

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the reaction between AgNO3 and NaBr to obtain AgBr and NaNO3. The balanced chemical equation is as follows: \[AgNO_{3}(aq) + NaBr(aq) \rightarrow AgBr(s) + NaNO_{3}(aq)\]

Find moles of each reactant

To find the limiting reactant, we must first determine the moles of each reactant present in the reaction. We will use the given volumes and concentrations to calculate the number of moles for each reactant. For AgNO3: Moles = (volume in L) × (concentration in M) Moles of AgNO3 = (100.0 mL × (1 L/1000 mL)) × (0.150 mol/L) Moles of AgNO3 = 0.015 mol For NaBr: Moles = (volume in L) × (concentration in M) Moles of NaBr = (20.0 mL × (1 L/1000 mL)) × (1.00 mol/L) Moles of NaBr = 0.020 mol

Identify the limiting reactant

Now, we will compare the moles of reactants to determine the limiting reactant. The stoichiometry of the balanced chemical equation shows us that 1 mole of AgNO3 reacts with 1 mole of NaBr. From the calculated moles of reactants: 0.015 mol AgNO3 × (1 mol NaBr / 1 mol AgNO3) = 0.015 mol NaBr Since we have 0.020 mol of NaBr, but only 0.015 mol are required to react with the available AgNO3, AgNO3 is the limiting reactant. The reaction will stop when all 0.015 mol of AgNO3 have been consumed.

Calculate the mass of AgBr produced

Since AgNO3 is the limiting reactant, the amount of AgBr produced will depend on the moles of AgNO3. We will use the stoichiometry of the balanced chemical equation to find the moles of AgBr produced. Moles of AgBr produced = (moles of limiting reactant AgNO3) × (1 mol AgBr / 1 mol AgNO3) Moles of AgBr produced = 0.015 mol Now, we will convert moles of AgBr to mass. To do this, we need to know the molar mass of AgBr, which is approximately 187.77 g/mol. Mass of AgBr = (moles of AgBr) × (molar mass of AgBr) Mass of AgBr = (0.015 mol) × (187.77 g/mol) Mass of AgBr = 2.82 g The mass of solid AgBr produced is approximately 2.82 g.

Key concepts

Limiting Reactant

Understanding the concept of a limiting reactant is crucial in stoichiometry. When two or more reactants participate in a chemical reaction, the limiting reactant is the substance that is completely consumed first, thus limiting the extent of the reaction and determining the amount of product formed. This means that even if other reactants are present in excess, the reaction can proceed only until the limiting reactant is used up.

In our original exercise, we found that 0.015 moles of AgNO3 and 0.020 moles of NaBr were available. According to the balanced chemical equation, the reaction between AgNO3 and NaBr occurs in a 1:1 molar ratio:

• 1 mole of AgNO3 reacts with 1 mole of NaBr.

Since 0.015 moles of AgNO3 require only 0.015 moles of NaBr to react completely, AgNO3 becomes the limiting reactant. This is because there is less of it in comparison to what would be necessary to fully consume the available NaBr. As a result, the amount of AgBr produced is governed by the amount of AgNO3 present.

Molar Mass Calculation

Calculating molar mass is an essential skill in chemistry, as it allows us to convert between the mass of a substance and the number of moles. The molar mass is the mass of one mole of a substance, typically expressed in grams per mole (\( ext{g/mol}\).

To determine the mass of AgBr produced in the original exercise, we first calculated the moles of AgBr, which were equivalent to the moles of our limiting reactant, AgNO3. To convert these moles into mass, we needed the molar mass of AgBr. To calculate this, we add the atomic masses of silver (Ag) and bromine (Br), which are roughly 107.87 g/mol and 79.90 g/mol, respectively:

• Molar mass of AgBr = ATOMIC MASS of Ag + ATOMIC MASS of Br = 107.87 + 79.90 = 187.77 g/mol

Thus, the molar mass of AgBr totals approximately 187.77 g/mol. Multiplying the moles of AgBr by this molar mass gives us the mass of AgBr formed in the reaction:

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. A balanced chemical equation reflects this transformation, showing the specific ratios in which reactants combine and products form. This balance is vital, as it follows the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.

In the original exercise, the reaction between AgNO3 and NaBr is described by the balanced equation:\[AgNO_{3(aq)} + NaBr_{(aq)} \rightarrow AgBr_{(s)} + NaNO_{3(aq)}\]This indicates that one molecule of AgNO3 reacts with one molecule of NaBr to produce one molecule of AgBr and one of NaNO3. The coefficients in this equation highlight the stoichiometric relationship, ensuring that the same number of each type of atom is present on both sides.

• This specific reaction is a type of double displacement, where the anions and cations of two different molecules switch places, forming two new compounds.

Through understanding the principles of chemical reactions, we can predict the quantities of products formed from various reactants, given their initial amounts.