Question

Consider the reaction between 0.156 L of \(0.105 M\) magnesium nitrate and 0.166 L of 0.106 \(M\) potassium hydroxide. a. What mass of precipitate will form? b. What is the concentration of nitrate ions left in solution after the reaction is complete?

Short answer

a. The mass of the precipitate formed is 0.81185 g. b. The concentration of nitrate ions left in the solution is 0.02355 M.

Step-by-step solution

Write the balanced chemical equation

The reaction between magnesium nitrate and potassium hydroxide is: \(Mg(NO_3)_2(aq) + 2 KOH(aq) \rightarrow Mg(OH)_2(s) + 2 KNO_3(aq)\)

Determine the limiting reactant

To determine the limiting reactant, we need to first calculate the moles of each reactant present in the solution. Moles of magnesium nitrate = Volume × Concentration Moles of magnesium nitrate = 0.156 L × 0.105 M = 0.01638 mol Moles of potassium hydroxide = Volume × Concentration Moles of potassium hydroxide = 0.166 L × 0.106 M = 0.017596 mol Now, determine the mole ratio for magnesium nitrate and potassium hydroxide from the balanced equation: \(1\:mol Mg(NO_3)_2 \leftrightarrow 2\: mol KOH\) We have 0.01638 mol of magnesium nitrate and 0.017596 mol of potassium hydroxide. Based on the stoichiometry, we need 2 times more moles of potassium hydroxide than magnesium nitrate. 0.01638 mol × 2 = 0.03276 mol of potassium hydroxide is needed. Since we have only 0.017596 mol of potassium hydroxide, the limiting reactant is potassium hydroxide.

Calculate the mass of precipitate formed

The mass of magnesium hydroxide (the precipitate) can be calculated using the stoichiometry of the balanced chemical equation and the molar mass. For each mole of potassium hydroxide, 0.5 mole of magnesium hydroxide is produced. Therefore, the moles of magnesium hydroxide produced: 0.017596 mol × 0.5 = 0.008798 mol of magnesium hydroxide Now, calculate the mass of magnesium hydroxide formed: Mass = Moles × Molar mass of magnesium hydroxide The molar mass of Mg(OH)₂ is 58.3 (Mg) + 2(15.999 (O) + 1.008 (H)) = 58.3 + 34.014 = 92.314 g/mol Mass = 0.008798 mol × 92.314 g/mol = 0.81185 g The mass of the precipitate formed is 0.81185 g.

Calculate the concentration of nitrate ions left in the solution

To find the concentration of nitrate ions remaining in the solution, we first need to calculate the moles of magnesium nitrate remaining in the solution after the reaction. Moles of Mg(NO₃)₂ used in the reaction with the limiting reactant (KOH): 0.008798 mol (as calculated above) Moles of magnesium nitrate remaining = Initial moles - Moles used in the reaction Moles of magnesium nitrate remaining = 0.01638 mol - 0.008798 mol = 0.007582 mol Now, calculate the total volume of the solution: Total volume = volume of magnesium nitrate solution + volume of potassium hydroxide solution Total volume = 0.156 L + 0.166 L = 0.322 L Finally, determine the concentration of the remaining nitrate ions: Concentration of nitrate ions = Moles of magnesium nitrate remaining / Total volume Concentration of nitrate ions = 0.007582 mol / 0.322 L = 0.02355 M The concentration of nitrate ions left in the solution is 0.02355 M. Answer: a. The mass of the precipitate formed is 0.81185 g. b. The concentration of nitrate ions left in the solution is 0.02355 M.

Key concepts

Limiting Reactant

Understanding the concept of a limiting reactant is crucial when dealing with chemical reactions. It can be compared to a recipe—if you have only two eggs but the recipe calls for three to make a cake, the number of cakes you can bake is limited by the number of eggs you have. Similarly, in chemistry, the limiting reactant is the substance that is completely consumed first in a chemical reaction. Once it's used up, the reaction stops, and no additional product can form.

In the given exercise, identifying the limiting reactant is a key step for predicting how much product will be formed. This is done by comparing the moles of the reactants to the proportions needed according to the balanced chemical equation. Once the limiting reactant is identified, we calculate how much of the product can be formed. This is inherently linked to the understanding of stoichiometry, that is, the quantitative relationship between reactants and products in a chemical reaction.

Molar Mass

The molar mass is another fundamental concept in chemistry, pivotal for various calculations involving chemical substances. To put it simply, molar mass is the weight of one mole of a substance, usually expressed in grams per mole (g/mol). The term 'mole' represents Avogadro's number of particles (atoms, molecules, ions) of that substance, which is approximately 6.022 x 10²³.

Calculating the molar mass involves adding up the atomic weights of all atoms in a molecule, as seen with magnesium hydroxide in the exercise. Here, the molar mass is used to convert moles of magnesium hydroxide into grams. This conversion is critical when you want to relate the mass of a compound produced in a lab to the theoretical calculation.

Precipitation Reaction

A precipitation reaction is a type of chemical reaction in which two soluble salts react in solution to form one or more insoluble products, referred to as precipitates. These are the solids that ‘fall out’ of the solution, and they can be collected by filtration.

In our exercise, when magnesium nitrate reacts with potassium hydroxide in aqueous solution, magnesium hydroxide, an insoluble compound, is formed as a precipitate. The solid nature of the precipitate in contrast to the remaining aqueous ions in solution, allows it to be easily separated and weighed, which directly applies to the determination of the mass of the product in the solution. The precipitation reaction concept is useful not only for making new compounds but also for removing unwanted substances from a liquid in processes such as water purification.