Question

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

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 \(2.82~g\).

Step-by-step solution

Write balanced chemical equation

First, we need a balanced chemical equation for the reaction between AgNO3 and NaBr. The chemical equation for the reaction is: \[ AgNO_3(aq) + NaBr(aq) \rightarrow AgBr(s) + NaNO_3(aq) \] This equation is already balanced.

Find moles of reactants

Now we need to calculate the moles of AgNO3 and NaBr, using their respective concentrations and volumes. Moles of AgNO3 = concentration × volume Moles of AgNO3 = \(0.150~M \times 100.0~mL\) Convert volume from mL to L: Moles of AgNO3 = \(0.150~M \times 0.100~L = 0.0150~mol\) Moles of NaBr = concentration × volume Moles of NaBr = \(1.00~M \times 20.0~mL\) Convert volume from mL to L: Moles of NaBr = \(1.00~M \times 0.0200~L = 0.0200~mol\)

Identify limiting reactant

To identify the limiting reactant, we need to compare the mole ratio of AgNO3 to NaBr with their stoichiometric ratio (1:1). Mole ratio = moles of AgNO3/moles of NaBr = \(0.0150~mol/0.0200~mol = 0.75\) Since the mole ratio is less than 1, AgNO3 is the limiting reactant. The reaction will stop when all AgNO3 is consumed.

Calculate moles of AgBr produced

Using the stoichiometric ratio from the balanced chemical equation (1:1), we can calculate the moles of AgBr formed: Moles of AgBr = moles of limiting reactant (AgNO3) = 0.0150 mol

Calculate mass of AgBr produced

Finally, we need to convert the moles of AgBr into mass, using the molar mass of AgBr (187.77 g/mol): Mass of AgBr = moles of AgBr × molar mass of AgBr Mass of AgBr = \(0.0150~mol \times 187.77~g/mol = 2.82~g\) Thus, 2.82 grams of solid AgBr is produced when 100.0 mL of 0.150 M AgNO3 is added to 20.0 mL of 1.00 M NaBr.

Key concepts

Stoichiometry

Stoichiometry is like a recipe for chemistry. It involves using the balanced chemical equations to figure out how much of each reactant you need and how much product you can make. It's kind of like cooking, where you need to know the right amounts of ingredients to end up with a delicious meal. In stoichiometry, instead of cups or tablespoons, we use moles, which tell us how many particles (like atoms or molecules) we have.

In the exercise with the silver nitrate (AgNO3) and sodium bromide (NaBr), stoichiometry helps us understand that when we mix these two chemicals, they react in a definite ratio to produce silver bromide (AgBr) and sodium nitrate (NaNO3). The balanced equation tells us that they react one-to-one: for each mole of AgNO3, we need one mole of NaBr to react completely. From this, educators can illustrate the importance of measuring ingredients in the lab accurately, just like you would with a cake recipe!

Balanced Chemical Equations

Now, a balanced chemical equation is crucial. It's like making sure you have the right number of socks to match your shoes. In the case of our reaction, the magic equation is:
\[ AgNO_3(aq) + NaBr(aq) \rightarrow AgBr(s) + NaNO_3(aq) \]
This tells us that one silver ion from AgNO3 bonds with one bromide ion from NaBr to form one molecule of AgBr, and the leftover sodium and nitrate ions pair up to make NaNO3. The equation is already balanced, meaning we have the same number of each element on both sides. This balance is a law in chemistry – it's like the universe's strict accounting rule: what you end with must equal what you started with. Without a balanced equation, you wouldn't be able to do stoichiometry correctly because you wouldn't know the precise 'recipe' the chemicals follow when they react.

Molar Mass Calculations

Molar mass is the weight of one mole of a substance, essentially a chemical compound's 'gram formula mass'. It's the total mass of that compound's atoms added up, and it's usually listed on the periodic table for each element (like weights in a grocery store). For our friend AgBr, this mass is found by adding up the mass of silver (Ag) and bromine (Br), which gives us its molar mass of 187.77 grams per mole.

To find out how much product (AgBr) we are getting from our reaction, we need to do two things: use stoichiometry to figure out the number of moles of AgBr we can make (based on the balanced equation and the limiting reactant), and then multiply by the molar mass of AgBr to convert moles to grams. It's like if someone told you how many batches of cookies you could make and you multiplied by the number of cookies per batch to get the total number of cookies. Simply, if you want to bake a perfect cake (or in this case, produce a certain chemical), you can't mess up the measurements!