Question

A solution of potassium chromate reacts with a solution of lead(II) nitrate to produce a yellow precipitate of lead(II) chromate and a solution of potassium nitrate. a. Write the balanced chemical equation. b. Starting with 0.250 mol of potassium chromate, determine the mass of lead chromate formed.

Short answer

The balanced chemical equation for the reaction is: $2K_{2}Cr{O}_4(aq)+Pb(N{O}_3{)}_2(aq)\to PbCr{O}_4(s)+2KN{O}_3(aq)$. When starting with 0.250 mol of potassium chromate, 40.4 g of lead(II) chromate are formed.

Step-by-step solution

1. Write the unbalanced equation

Write the equation with the correct chemical formulas for each reactant and product: $K_{2}Cr{O}_4(aq)+Pb(N{O}_3{)}_2(aq)\to PbCr{O}_4(s)+KN{O}_3(aq)$

2. Balance the equation

To balance the equation, we need to make sure the number of atoms is equal on both sides: $2K_{2}Cr{O}_4(aq)+Pb(N{O}_3{)}_2(aq)\to PbCr{O}_4(s)+2KN{O}_3(aq)$ Now the equation is balanced with 2 moles of potassium chromate reacting with 1 mole of lead nitrate to produce 1 mole of lead chromate and 2 moles of potassium nitrate. #b. Determine the mass of lead(II) chromate formed#

3. Calculate moles of lead(II) chromate using stoichiometry

Given that we start with 0.250 mol of potassium chromate, we can use the balanced chemical equation to determine the moles of lead(II) chromate formed: $$2moles{K}_{2}Cr{O}_4\to 1molePbCr{O}_4$$ Therefore, the number of moles of lead(II) chromate: $0.250mole{K}_{2}Cr{O}_4\times \frac{1molePbCr{O}_4}{2moles{K}_{2}Cr{O}_4}=0.125molePbCr{O}_4$

4. Calculate the mass of lead(II) chromate

Now that we have the number of moles, we can determine the mass of lead(II) chromate. First, find the molar mass of lead(II) chromate: $$MolarmassofPbCr{O}_4=207.2(\mathrm{Pb})+51.996(\mathrm{Cr})+64(\mathrm{O})=323.2g/mol$$ Finally, multiply the moles of lead(II) chromate by its molar mass to find its mass in grams: $0.125molePbCr{O}_4\times \frac{323.2g}{1molePbCr{O}_4}=40.4g$ The mass of lead(II) chromate formed is 40.4 g.

Key concepts

Balanced Chemical Equation

In stoichiometry, balancing a chemical equation is as fundamental as ensuring you have the correct ingredients in a recipe. It requires accounting for all the atoms of the reactants and products involved in a chemical reaction. Let's take the chemical reaction where potassium chromate reacts with lead(II) nitrate as an example.

When you first write the chemical formula, it often won't be balanced. With our example, the chemical equation starts as an unbalanced one with the formulas: $K_{2}Cr{O}_4(aq)+Pb(N{O}_3{)}_2(aq)\to PbCr{O}_4(s)+KN{O}_3(aq)$. To balance this, we adjust the coefficients to ensure that the number of each type of atom on the reactants side is equal to the number on the products side. This gives us the balanced equation: $2K_{2}Cr{O}_4(aq)+Pb(N{O}_3{)}_2(aq)\to PbCr{O}_4(s)+2KN{O}_3(aq)$.

A crucial insight is that these coefficients represent the moles of each substance involved in the reaction. A balanced equation provides the mole ratio needed for the stoichiometric calculations that are essential for predicting the quantities of substances produced or consumed in the reaction.

Precipitate Formation

Precipitate formation is a fascinating outcome of certain chemical reactions, especially when two soluble salts react in solution to form an insoluble compound. This is exactly what happens in the reaction between potassium chromate and lead(II) nitrate. The yellow precipitate you see is lead(II) chromate, a solid that forms out of the aqueous mixture.

A precipitate forms when the product of the reaction is insoluble in the solvent, in this case water. It's the result of ionic compounds reacting and the ions combining to form a new compound that is not soluble. The formation of a precipitate can be indicative of a chemical change and is used in various applications, from water treatment to qualitative chemical analysis. Recognizing a precipitate is also important in laboratory settings to confirm the occurrence of certain reactions.

Mole-to-Mass Calculation

Understanding mole-to-mass calculations is paramount in stoichiometry for converting between the mole (the chemist’s counting unit) and mass (something we can physically measure). Once you have the balanced chemical equation, you can use the mole ratios it provides to perform these conversions.

Let's use the current problem as our learning context. We're trying to figure out how much lead(II) chromate is produced from 0.250 moles of potassium chromate. The mole ratio from the balanced equation informs us that 2 moles of potassium chromate yields 1 mole of lead(II) chromate. We set up a proportion: $0.250\text{ moles }{K}_{2}Cr{O}_4\times \frac{1\text{ mole }PbCr{O}_4}{2\text{ moles }{K}_{2}Cr{O}_4}=0.125\text{ moles }PbCr{O}_4$.

The next step is to convert moles of lead(II) chromate to mass. To do this, we need the molar mass, which for lead(II) chromate is 323.2 g/mol. Multiplying the moles by the molar mass, $0.125\text{ moles }PbCr{O}_4\times \frac{323.2\text{ g}}{1\text{ mole }PbCr{O}_4}=40.4\text{ g}$, gives us the mass. Thus, the beauty of stoichiometry shines in transforming abstractions (moles) into tangible quantities (grams) we can measure.