Question

A sample of 7.75 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?

Short answer

The balanced chemical equation for the reaction between magnesium hydroxide (Mg(OH)₂) and nitric acid (HNO₃) is: $$Mg(OH{)}_2+2HN{O}_3\to Mg(N{O}_3{)}_2+2H_{2}O$$ The limiting reactant in the reaction is HNO₃. After the reaction is complete, there are 0.1304 moles of Mg(OH)₂, 0 moles of HNO₃, and 0.0025 moles of Mg(NO₃)₂ present.

Step-by-step solution

1. Write the balanced chemical equation

The reaction between magnesium hydroxide (Mg(OH)₂) and nitric acid (HNO₃) is an acid-base reaction, forming magnesium nitrate (Mg(NO₃)₂) and water (H₂O) as products. The balanced chemical equation for the reaction is: $$Mg(OH{)}_2+2HN{O}_3\to Mg(N{O}_3{)}_2+2H_{2}O$$

2. Calculate the initial moles of both reactants

We are given: - Mass of Mg(OH)₂ = 7.75 g - Volume of HNO₃ solution = 25.0 mL - Concentration of HNO₃ solution = 0.200 M Now let's calculate the initial moles of Mg(OH)₂ and HNO₃. (a) Moles of Mg(OH)₂: To calculate moles, we use the formula: $$moles=\frac{mass}{molarmass}$$ The molar mass of Mg(OH)₂ is: (24.3 g/mol for Mg) + (2 × (16.0 g/mol for O) + (1.01 g/mol for H)) = 58.3 g/mol So, the moles of Mg(OH)₂: $$molesofMg(OH{)}_2=\frac{7.75g}{58.3g/mol}=0.1329mol$$ (b) Moles of HNO₃: Since we know the concentration and volume of the HNO₃ solution, we can calculate the moles using the formula: $$moles=concentration\times volume(inliters)$$ The volume of the HNO₃ solution in liters is: $25.0mL=\frac{25.0}{1000}L=0.025L$ Now, let's calculate the moles of HNO₃: $$molesofHN{O}_3=0.200M\times 0.025L=0.005mol$$

3. Determine the limiting reactant

In the balanced chemical equation: $$Mg(OH{)}_2+2HN{O}_3\to Mg(N{O}_3{)}_2+2H_{2}O$$ The stoichiometry shows that 1 mole of Mg(OH)₂ reacts with 2 moles of HNO₃. We will divide the initial moles of each reactant by its stoichiometric coefficient to determine the limiting reactant: Mg(OH)₂: $\frac{0.1329mol}{1}=0.1329mol$ HNO₃: $\frac{0.005mol}{2}=0.0025mol$ Since 0.0025 mol of HNO₃ is less than 0.1329 mol of Mg(OH)₂, HNO₃ is the limiting reactant.

4. Calculate the moles of all species after the reaction is complete

Since HNO₃ is the limiting reactant, all 0.005 mol of it will be used up in the reaction. The stoichiometry of the reaction shows that 1 mole of Mg(OH)₂ reacts with 2 moles of HNO₃, so the change in moles of Mg(OH)₂ is: $$\mathrm{\Delta }molofMg(OH{)}_2=-\frac{0.005mol}{2}=-0.0025mol$$ For Mg(NO₃)₂ and H₂O, we apply the stoichiometry from the balanced equation: $$Mg(OH{)}_2+2HN{O}_3\to Mg(N{O}_3{)}_2+2H_{2}O$$ $$0.0025mol0.0050mol$$ So, the moles of Mg(OH)₂, HNO₃, and Mg(NO₃)₂ after the reaction is complete are: Mg(OH)₂: $0.1329mol-0.0025mol=0.1304mol$ HNO₃: $0.005mol-0.005mol=0mol$ Mg(NO₃)₂: $0+0.0025mol=0.0025mol$ Thus, after the reaction is complete, there are 0.1304 moles of Mg(OH)₂, 0 moles of HNO₃, and 0.0025 moles of Mg(NO₃)₂ present.

Key concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the reactant that runs out first. This happens because it is consumed completely while the other reactants may still be left in excess. The quantity of the limiting reactant determines the amount of products that can be formed.

To identify the limiting reactant, we need to use the balanced chemical equation and the initial amount of each reactant. In this exercise, we deal with magnesium hydroxide, ${\mathrm{Mg}(\mathrm{OH})}_2$, and nitric acid, ${\mathrm{HNO}}_3$. The balanced chemical equation is:

• ${\mathrm{Mg}(\mathrm{OH})}_2+2{\mathrm{HNO}}_3\to {\mathrm{Mg}(\mathrm{NO}}_3{)}_2+2{\mathrm{H}}_2\mathrm{O}$

For every 1 mole of ${\mathrm{Mg}(\mathrm{OH})}_2$, 2 moles of ${\mathrm{HNO}}_3$ are required. By calculating the moles of both reactants and comparing them using the stoichiometry, we can find the limiting reactant.

Given that the reaction requires two moles of ${\mathrm{HNO}}_3$ for every mole of ${\mathrm{Mg}(\mathrm{OH})}_2$, a quick calculation shows that the amount of ${\mathrm{HNO}}_3$ present (0.005 mol) is not enough to react with all the ${\mathrm{Mg}(\mathrm{OH})}_2$ supplied (0.1329 mol). Thus, ${\mathrm{HNO}}_3$ is the limiting reactant.

Acid-Base Reaction

An acid-base reaction is a chemical reaction that occurs between an acid and a base. In our exercise, ${\mathrm{HNO}}_3$, a known strong acid, reacts with ${\mathrm{Mg}(\mathrm{OH})}_2$, which acts as a base. The reaction results in the formation of magnesium nitrate ${\mathrm{Mg}(\mathrm{NO}}_3{)}_2$ and water ${\mathrm{H}}_2\mathrm{O}$.

These reactions are always characterized by the transfer of a proton (${\mathrm{H}}^{+}$) from the acid to the base. Here’s how this reaction can be understood:

• The nitric acid ${\mathrm{HNO}}_3$ donates ${\mathrm{H}}^{+}$ ions.
• The magnesium hydroxide ${\mathrm{Mg}(\mathrm{OH})}_2$, a base, accepts the ${\mathrm{H}}^{+}$ ions.
• This results in the formation of water molecules and a salt, in this case, magnesium nitrate.

Understanding these reactions is crucial because they form the basis of several processes, including titrations in analytical chemistry, and the production of various compounds in industrial reactions.

Balanced Chemical Equation

A balanced chemical equation is essential for understanding chemical reactions because it provides the correct proportionate amount of each reactant and product involved in the reaction.

In this case, balancing the equation helps us know how many moles of ${\mathrm{Mg}(\mathrm{OH})}_2$ and ${\mathrm{HNO}}_3$ are needed, and what products we get. The equation is:

• ${\mathrm{Mg}(\mathrm{OH})}_2+2{\mathrm{HNO}}_3\to {\mathrm{Mg}(\mathrm{NO}}_3{)}_2+2{\mathrm{H}}_2\mathrm{O}$

This equation is balanced because the number of atoms of each element is the same on both sides.

• 1 magnesium (Mg) atom is on both sides.
• 2 oxygen (O) atoms in ${\mathrm{Mg}(\mathrm{OH})}_2$ and 4 in 2${\mathrm{HNO}}_3$, totalling 6 oxygens match perfectly with 4 in ${\mathrm{Mg}(\mathrm{NO}}_3{)}_2$ and 2 in ${\mathrm{H}}_2\mathrm{O}$.
• 4 hydrogen (H) atoms from 2${\mathrm{HNO}}_3$ are balanced by 4 hydrogen atoms in 2 ${\mathrm{H}}_2\mathrm{O}$.

Balancing is vital because it reflects the law of conservation of mass and ensures that calculations such as those for limiting reactants are accurate.