Problem 109 - Chemistry: The Central Science

Question

A sample of \(7.75 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is added to \(25.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3}\). (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) How many moles of \(\mathrm{Mg}(\mathrm{OH})_{2}, \mathrm{HNO}_{3}\), and \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) are present after the reaction is complete?

Step-by-Step Solution

Answer

In conclusion, after the reaction is complete, there are 0.1304 moles of Mg(OH)_2, 0 moles of HNO_3, and 0.0025 moles of Mg(NO_3)_2 present.

Step 1: Write the balanced chemical equation

To write the balanced chemical equation, we need to know the products formed when magnesium hydroxide reacts with nitric acid. Magnesium hydroxide is a base and nitric acid is an acid, so they form a salt (magnesium nitrate) and water: \[\mathrm{Mg(OH)_2 + 2HNO_3 \rightarrow Mg(NO_3)_2 + 2H_2O}\]

Step 2: Calculate moles of Mg(OH)_2 and HNO_3

Next, we need to calculate the number of moles of magnesium hydroxide and nitric acid present initially. Given mass of Mg(OH)_2 = 7.75 g Molar mass of Mg(OH)_2 = 58.3 g/mol (24.3 g/mol for Mg + 2 * (16.0 g/mol for O + 1.0 g/mol for H)) Moles of Mg(OH)_2 = (7.75 g) / (58.3 g/mol) = 0.1329 mol Given volume of 0.200 M HNO_3 = 25.0 mL Moles of HNO_3 = (25.0 mL) * (1 L / 1000 mL) * (0.200 mol/L) = 0.005 mol

Step 3: Determine the limiting reactant

Now we need to determine the limiting reactant by comparing the mole ratios of the reactants based on the balanced chemical equation: Mole ratio of Mg(OH)_2 to HNO_3 = 1:2 Moles of HNO_3 needed to react with 0.1329 mol of Mg(OH)_2 = 0.1329 mol * 2 = 0.2658 mol Since we only have 0.005 mol of HNO_3, which is less than 0.2658 mol needed to completely react with Mg(OH)_2, HNO_3 is the limiting reactant.

Step 4: Calculate moles of Mg(OH)_2, HNO_3, and Mg(NO_3)_2 after the reaction

To determine the moles of Mg(OH)_2, HNO_3, and Mg(NO_3)_2 after the reaction, we need to use the stoichiometry from the balanced chemical equation and the limiting reactant: Moles of HNO_3 consumed = 0.005 mol Moles of Mg(OH)_2 consumed = 0.005 mol / 2 = 0.0025 mol Moles of Mg(NO_3)_2 formed = 0.0025 mol (since the ratio of Mg(OH)_2 to Mg(NO_3)_2 is 1:1) After the reaction: Remaining moles of Mg(OH)_2 = 0.1329 mol - 0.0025 mol = 0.1304 mol Remaining moles of HNO_3 = 0.005 mol - 0.005 mol = 0 mol Moles of Mg(NO_3)_2 = 0.0025 mol In conclusion, after the reaction is complete, there are 0.1304 moles of Mg(OH)_2, 0 moles of HNO_3, and 0.0025 moles of Mg(NO_3)_2 present.

Key Concepts

Limiting Reactant

Understanding the concept of the limiting reactant is crucial to mastering stoichiometry. The limiting reactant in a chemical reaction is the substance that is completely consumed first, thus determining the amount of product that can be formed. This is because once the limiting reactant is used up, the reaction stops, no matter how much of the other reactants is left.

In the given example, the reactants are magnesium hydroxide (\(Mg(OH)_2\)) and nitric acid (\(HNO_3\)). To identify the limiting reactant, we first need to know the reactants' mole ratio in the balanced chemical equation. Then, by comparing the actual available moles of reactants to the mole ratio, we can determine which reactant will run out first.

For the exercise improvement, it's helpful to visualize the concept by thinking of the reactants as ingredients in a recipe. If a cake recipe requires two eggs per cake and you have ten eggs, how many cakes can you make if you only have enough flour for five cakes? The flour in this case would be the limiting 'reactant,' and the potential number of cakes ('products') is determined by it, not the eggs.

Chemical Equation

A chemical equation is a symbolic representation of a chemical reaction, showing the reactants and the products with their respective quantities. Balancing a chemical equation is necessary so that it obeys the law of conservation of mass, meaning the number of atoms for each element must be the same on both the reactants and products side.

The balanced chemical equation for the given exercise is \[Mg(OH)_2 + 2HNO_3 \rightarrow Mg(NO_3)_2 + 2H_2O\]. This equation tells us that one mole of magnesium hydroxide reacts with two moles of nitric acid to produce one mole of magnesium nitrate and two moles of water. During the exercise improvement process, emphasizing the importance of the coefficients in front of each substance is essential. These coefficients transform a qualitative description into a quantitative formula that can be used to calculate the amounts of reactants and products involved in the reaction.

Mole Calculation

The mole is a fundamental unit in chemistry that measures the amount of substance. Mole calculations provide a bridge between the mass of a substance and the number of particles or units of that substance. In stoichiometry, mole calculations are used to convert between mass and moles and to relate amounts of reactants and products through the coefficients in a balanced chemical equation.

To solve the provided exercise, we must perform mole calculations using the molar mass of magnesium hydroxide and the volumetric concentration of nitric acid. Once we calculate how many moles of each reactant we have, we can use stoichiometry to predict the results of the chemical reaction. For exercise improvement, emphasizing the relationship between grams, moles, and molar masses, as well as the concept of molarity (moles per liter), could greatly aid in understanding. These calculations allow us to determine the amount of product that will form and also indicate how much of each reactant remains after the reaction.