Precipitates and Calculations - AP Chemistry

First, let us write out an ion exchance reaction for the reactants:

Na_$2SO_{4hspace{1 mm}$$(aq)}+BaCl_{2hspace{1 mm}$(aq)}rightarrow 2NaClhspace{1 mm}+BaSO_4

By solubility rules, NaCl is soluble in water and BaSO_4 is not. Our new reaction is:

Na_$2SO_{4hspace{1 mm}$$(aq)}+BaCl_{2hspace{1 mm}$(aq)}rightarrow $2NaCl_{(aq)}$hspace{1 $mm}+BaSO_{4hspace{1 mm}$(s)}

Now we will calculate the theoretical yield of each reactant.

300hspace{1 mm}mLtimes$\frac{1 L}{1000 mL}$times $\frac{0.3hspace{1 mm}$moleshspace{1 mm}Na_2SO_4}{1 L}times$\frac{1hspace{1 mm}$molehspace{1 mm}BaSO_4}{1hspace{1 mm}molehspace{1 mm}Na_2SO_4}times $\frac{233.4hspace{1 mm}$ghspace{1 mm}BaSO_4}{1hspace{1 mm}molehspace{1 mm}BaSO_4}=21.0hspace{1 mm}ghspace{1 mm}BaSO_4

Now we perform the same calculation beginning with BaCl_2:

200hspace{1 mm}mLtimes$\frac{1hspace{1 mm}$L}{1000hspace{1 mm}mL}times$\frac{0.4hspace{1 mm}$moleshspace{1 mm}BaCl_2}{1hspace{1 mm}L}times$\frac{1hspace{1 mm}$molehspace{1 mm}BaSO_4}{1hspace{1 mm}molehspace{1 mm}BaCl_2}times$\frac{233.4hspace{1 mm}$ghspace{1 mm}BaSO_4}{1hspace{1 mm}molehspace{1 mm}BaSO_4}=18.7hspace{1 mm}ghspace{1 mm}BaSO_4

The limiting reagent is BaCl_2 and this reaction produces 18.7 g precipitate.

234hspace{1 mm}mLtimes$\frac{1hspace{1 mm}$L}{1000hspace{1 mm}mL}times$\frac{0.4hspace{1 mm}$moleshspace{1 mm}KCl}{1hspace{1 mm}L}times$\frac{74.55hspace{1 mm}$ghspace{1 mm}KCl}{1hspace{1 mm}molehspace{1 mm}KCl}=6.98hspace{1 mm}ghspace{1 mm}KCl

First, you must recognize that the chemical formula for sodium hydroxide is . The mass of the boiled solution is

The remaining mass with be equal to the mass of the sodium chloride in the solution. Once the solvent (water) evaporates, the solute will remain.

Atomic mass of sodium is ~23. Atomic mass of chlorine is ~35.5. Molecular mass of NaCl is ~58.5.

The dissociation of calcium hydroxide in aqueous solution is:

The of calcium hydroxide is related to the dissolved concentrations of its counterions:

and are produced in a molar ratio of 1:2; for each molecule of calcium hydroxide that dissolves:

Given a value of , the molar solubilities of each counterion may be determined by setting . It follows that:

Now, we can use basic algebra to solve for :

Since we set , and , multiplying the value of by two gives the correct answer, which is 0.022M.

Use the Henderson-Hasselbalch equation to determine the ratio of bicarbonate to carbonic acid in solution:

Solve for the ratio we need to answer the question:

Double replacement reactions can be further categorized as precipitation reactions since it is possible to make a precipitate (solid) from mixing two liquids. Combustion reactions involve using oxygen to burn another species, and the products are carbon dioxide and water. Combination reactions involve the synthesis of one molecule from two separate ones; decomposition is the opposite.

234hspace{1 mm}mLtimes$\frac{1hspace{1 mm}$L}{1000hspace{1 mm}mL}times$\frac{0.4hspace{1 mm}$moleshspace{1 mm}KCl}{1hspace{1 mm}L}times$\frac{74.55hspace{1 mm}$ghspace{1 mm}KCl}{1hspace{1 mm}molehspace{1 mm}KCl}=6.98hspace{1 mm}ghspace{1 mm}KCl

First, you must recognize that the chemical formula for sodium hydroxide is . The mass of the boiled solution is

First we will figure out the number of moles of Li_3 PO_4 that we have. We have 5.2 L of a 0.3 M solution, so:

0.3hspace{1mm} $\frac{mol}{L}$ times 5.2hspace{1 mm} L = 1.56 hspace{1 mm}moles

Now the problem is to find the mass of 1.56 moles of Li_3 PO_4. First, we need to know the molecular weight of Li_3 PO_4. We will go to the periodic table and add up the mass of each element present.

(3hspace{1 mm}times6.94hspace{1 mm}$\frac{gramshspace{1 mm}$ Li}{molhspace{1 mm}Li})+30.97hspace{1 mm}$\frac{gramshspace{1 mm}$P}{molhspace{1 mm}P}+(4hspace{1 mm}times 16hspace{1 mm}$\frac{gramshspace{1 mm}$O}{molhspace{1mm}O})= 115.79hspace{1mm}$\frac{gramshspace{1 mm}$ Li_3 PO_4}{molhspace{1 mm}Li_3 PO_4}

Now the problem is simple as we have the molar mass and the number of desired moles.

115.79hspace{1 mm}$\frac{gramshspace{1 mm}$Li_3 PO_4}{molhspace{1 mm}Li_3 PO_4}times1.56hspace{1 mm}moleshspace{1 mm}Li_3 PO_4 = 181hspace{1 mm}gramshspace{1 mm}Li_3 PO_4

First, let us write out an ion exchance reaction for the reactants:

Na_$2SO_{4hspace{1 mm}$$(aq)}+BaCl_{2hspace{1 mm}$(aq)}rightarrow 2NaClhspace{1 mm}+BaSO_4

By solubility rules, NaCl is soluble in water and BaSO_4 is not. Our new reaction is:

Na_$2SO_{4hspace{1 mm}$$(aq)}+BaCl_{2hspace{1 mm}$(aq)}rightarrow $2NaCl_{(aq)}$hspace{1 $mm}+BaSO_{4hspace{1 mm}$(s)}

Now we will calculate the theoretical yield of each reactant.

300hspace{1 mm}mLtimes$\frac{1 L}{1000 mL}$times $\frac{0.3hspace{1 mm}$moleshspace{1 mm}Na_2SO_4}{1 L}times$\frac{1hspace{1 mm}$molehspace{1 mm}BaSO_4}{1hspace{1 mm}molehspace{1 mm}Na_2SO_4}times $\frac{233.4hspace{1 mm}$ghspace{1 mm}BaSO_4}{1hspace{1 mm}molehspace{1 mm}BaSO_4}=21.0hspace{1 mm}ghspace{1 mm}BaSO_4

Now we perform the same calculation beginning with BaCl_2:

200hspace{1 mm}mLtimes$\frac{1hspace{1 mm}$L}{1000hspace{1 mm}mL}times$\frac{0.4hspace{1 mm}$moleshspace{1 mm}BaCl_2}{1hspace{1 mm}L}times$\frac{1hspace{1 mm}$molehspace{1 mm}BaSO_4}{1hspace{1 mm}molehspace{1 mm}BaCl_2}times$\frac{233.4hspace{1 mm}$ghspace{1 mm}BaSO_4}{1hspace{1 mm}molehspace{1 mm}BaSO_4}=18.7hspace{1 mm}ghspace{1 mm}BaSO_4

The limiting reagent is BaCl_2 and this reaction produces 18.7 g precipitate.

The remaining mass with be equal to the mass of the sodium chloride in the solution. Once the solvent (water) evaporates, the solute will remain.

Atomic mass of sodium is ~23. Atomic mass of chlorine is ~35.5. Molecular mass of NaCl is ~58.5.

The dissociation of calcium hydroxide in aqueous solution is:

The of calcium hydroxide is related to the dissolved concentrations of its counterions:

and are produced in a molar ratio of 1:2; for each molecule of calcium hydroxide that dissolves:

Given a value of , the molar solubilities of each counterion may be determined by setting . It follows that:

Now, we can use basic algebra to solve for :

Since we set , and , multiplying the value of by two gives the correct answer, which is 0.022M.

Use the Henderson-Hasselbalch equation to determine the ratio of bicarbonate to carbonic acid in solution:

Solve for the ratio we need to answer the question:

Double replacement reactions can be further categorized as precipitation reactions since it is possible to make a precipitate (solid) from mixing two liquids. Combustion reactions involve using oxygen to burn another species, and the products are carbon dioxide and water. Combination reactions involve the synthesis of one molecule from two separate ones; decomposition is the opposite.

234hspace{1 mm}mLtimes$\frac{1hspace{1 mm}$L}{1000hspace{1 mm}mL}times$\frac{0.4hspace{1 mm}$moleshspace{1 mm}KCl}{1hspace{1 mm}L}times$\frac{74.55hspace{1 mm}$ghspace{1 mm}KCl}{1hspace{1 mm}molehspace{1 mm}KCl}=6.98hspace{1 mm}ghspace{1 mm}KCl

First, you must recognize that the chemical formula for sodium hydroxide is . The mass of the boiled solution is

First we will figure out the number of moles of Li_3 PO_4 that we have. We have 5.2 L of a 0.3 M solution, so:

0.3hspace{1mm} $\frac{mol}{L}$ times 5.2hspace{1 mm} L = 1.56 hspace{1 mm}moles

Now the problem is to find the mass of 1.56 moles of Li_3 PO_4. First, we need to know the molecular weight of Li_3 PO_4. We will go to the periodic table and add up the mass of each element present.

(3hspace{1 mm}times6.94hspace{1 mm}$\frac{gramshspace{1 mm}$ Li}{molhspace{1 mm}Li})+30.97hspace{1 mm}$\frac{gramshspace{1 mm}$P}{molhspace{1 mm}P}+(4hspace{1 mm}times 16hspace{1 mm}$\frac{gramshspace{1 mm}$O}{molhspace{1mm}O})= 115.79hspace{1mm}$\frac{gramshspace{1 mm}$ Li_3 PO_4}{molhspace{1 mm}Li_3 PO_4}

Now the problem is simple as we have the molar mass and the number of desired moles.

115.79hspace{1 mm}$\frac{gramshspace{1 mm}$Li_3 PO_4}{molhspace{1 mm}Li_3 PO_4}times1.56hspace{1 mm}moleshspace{1 mm}Li_3 PO_4 = 181hspace{1 mm}gramshspace{1 mm}Li_3 PO_4

First, let us write out an ion exchance reaction for the reactants:

Na_$2SO_{4hspace{1 mm}$$(aq)}+BaCl_{2hspace{1 mm}$(aq)}rightarrow 2NaClhspace{1 mm}+BaSO_4

By solubility rules, NaCl is soluble in water and BaSO_4 is not. Our new reaction is:

Na_$2SO_{4hspace{1 mm}$$(aq)}+BaCl_{2hspace{1 mm}$(aq)}rightarrow $2NaCl_{(aq)}$hspace{1 $mm}+BaSO_{4hspace{1 mm}$(s)}

Now we will calculate the theoretical yield of each reactant.

300hspace{1 mm}mLtimes$\frac{1 L}{1000 mL}$times $\frac{0.3hspace{1 mm}$moleshspace{1 mm}Na_2SO_4}{1 L}times$\frac{1hspace{1 mm}$molehspace{1 mm}BaSO_4}{1hspace{1 mm}molehspace{1 mm}Na_2SO_4}times $\frac{233.4hspace{1 mm}$ghspace{1 mm}BaSO_4}{1hspace{1 mm}molehspace{1 mm}BaSO_4}=21.0hspace{1 mm}ghspace{1 mm}BaSO_4

Now we perform the same calculation beginning with BaCl_2:

200hspace{1 mm}mLtimes$\frac{1hspace{1 mm}$L}{1000hspace{1 mm}mL}times$\frac{0.4hspace{1 mm}$moleshspace{1 mm}BaCl_2}{1hspace{1 mm}L}times$\frac{1hspace{1 mm}$molehspace{1 mm}BaSO_4}{1hspace{1 mm}molehspace{1 mm}BaCl_2}times$\frac{233.4hspace{1 mm}$ghspace{1 mm}BaSO_4}{1hspace{1 mm}molehspace{1 mm}BaSO_4}=18.7hspace{1 mm}ghspace{1 mm}BaSO_4

The limiting reagent is BaCl_2 and this reaction produces 18.7 g precipitate.