Question

A sample of $8.69\mathrm{g}$ of $\mathrm{Zn}(\mathrm{OH}{)}_2$ is added to $155.0\mathrm{mL}$ of $0.750\mathrm{M}{\mathrm{H}}_2{\mathrm{SO}}_4.$ (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) How many moles of $\mathrm{Zn}(\mathrm{OH}{)}_2,{\mathrm{H}}_2{\mathrm{SO}}_4,$ and ${\mathrm{ZnSO}}_4$ are present after the reaction is complete?

Short answer

The balanced chemical equation for the reaction is: $Zn(OH{)}_2+H_{2}S{O}_4\to ZnS{O}_4+2H_{2}O$. Zn(OH)₂ is the limiting reactant. After the reaction is complete, there are 0 moles of $Zn(OH{)}_2$, 0.0289 moles of $H_{2}S{O}_4$, and 0.0874 moles of $ZnS{O}_4$.

Step-by-step solution

Write the balanced chemical equation

First, we have to determine the products of the reaction between Zn(OH)₂ and H₂SO₄. Zn(OH)₂ is a metal hydroxide while H₂SO₄ is an acid, so based on the acid-base reaction, the products are salt and water. The balanced chemical equation: $Zn(OH{)}_2+H_{2}S{O}_4\to ZnS{O}_4+2H_{2}O$

Find the moles of given reactants

We need to find the number of moles of Zn(OH)₂ and H₂SO₄ in the given amounts, using the formula mass(molar_mass) and molarity. For Zn(OH)₂: The molar mass of Zn = 65.38 g/mol The molar mass of O = 16.00 g/mol The molar mass of H = 1.01 g/mol The formula mass of Zn(OH)₂ = (65.38 + 2 * (16.00 + 1.01)) = 65.38 + 2 * 17.01 = 99.40 g/mol From the given amount of Zn(OH)₂ (8.69 g), the number of moles = 8.69 g ÷ 99.40 g/mol = 0.0874 mol For H₂SO₄: The given molarity of H₂SO₄ = 0.750 M, which means 0.750 moles are present in 1 L of solution. So, in 155 mL (0.155 L), the number of moles =0.750 mol * 0.155 L = 0.1163 mol

Identify the limiting reactant

We can use the stoichiometry of the balanced chemical equation to calculate the required moles of H₂SO₄ for the given moles of Zn(OH)₂: $Requiredmolesof{H}_{2}S{O}_4=1\ast 0.0874molZn(OH{)}_2=0.0874mol{H}_{2}S{O}_4$ As we have 0.1163 mol of H₂SO₄, it is more than required, so Zn(OH)₂ is the limiting reactant.

Calculate the moles of reactants and products after the reaction

As Zn(OH)₂ is the limiting reactant, all of it will be consumed, so there will be 0 moles of Zn(OH)₂ left after the reaction. Moles of H₂SO₄ consumed = moles of Zn(OH)₂ = 0.0874 mol Remaining moles of H₂SO₄ = 0.1163 - 0.0874 = 0.0289 mol Based on the stoichiometry of the balanced equation, 1 mol of Zn(OH)₂ reacts with 1 mol of H₂SO₄ to produce 1 mol of ZnSO₄. So, the moles of ZnSO₄ produced = moles of Zn(OH)₂ consumed = 0.0874 mol In summary, after the reaction is complete, we have: - moles of Zn(OH)₂ = 0 mol - moles of H₂SO₄ = 0.0289 mol - moles of ZnSO₄ = 0.0874 mol

Key concepts

Acid-Base Reactions

An acid-base reaction occurs when an acid and a base interact to form water and a salt. This is a classic chemical reaction in which an acid donates a proton (H⁺) to a base. In the case of our reaction, zinc hydroxide, Zn(OH)₂, acts as the base, and sulfuric acid, H₂SO₄, is the acid.
The reaction results in the formation of zinc sulfate (a salt) and water. Understanding acid-base reactions helps us predict the products and the direction of chemical changes in many common industrial and biological processes. Acid-base reactions are essential for various chemical calculations and are one of the fundamental types of reactions in chemistry.

Balanced Chemical Equation

A balanced chemical equation accurately represents the quantities of reactants and products in a reaction. It reflects the conservation of mass, where the amount of atoms for each element is the same on both sides of the equation. In our reaction, Zn(OH)₂ and H₂SO₄ react to produce ZnSO₄ and water.
The balanced equation is written as:
$$Zn(OH{)}_2+H_{2}S{O}_4\to ZnS{O}_4+2H_{2}O$$
Balancing involves adjusting the coefficients so the number of atoms of each element is equal on both sides. This ensures that no matter is created or destroyed during the reaction. Balancing chemical equations is a crucial skill in chemistry as it is foundational for understanding stoichiometry and reaction dynamics.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It uses the balanced chemical equation to determine the relationships between the quantities of substances involved. In our reaction, stoichiometry was used to compare the actual moles of substances we have with those required according to the balanced equation.
Key steps involve:

• Identifying the mole ratio between the reactants and products from the balanced equation.
• Calculating quantities based on these ratios to predict how much of each substance will react or be produced.

Understanding stoichiometry is vital for predicting the outcomes of reactions, planning chemical syntheses, and economic use of reactants.

Moles Calculation

Calculating moles helps determine the amount of a substance in a given mass or volume. It is done using the formula:
$$\text{number of moles}=\frac{\text{mass of substance (g)}}{\text{molar mass (g/mol)}}$$
For solutions, it can also be calculated using molarity:
$$\text{moles}=\text{molarity (mol/L)}\times \text{volume (L)}$$
In our example, we calculated the moles of Zn(OH)₂ from its given mass and molar mass. Similarly, we found moles of H₂SO₄ using its molarity and the volume of the solution.
Calculating moles is fundamental to understanding chemical reactions, as it relates to the quantity of reactants used and products formed, forming the basis for further stoichiometric calculations.