Problem 109 - Chemistry The Central Science In Si Units

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?

Step-by-Step Solution

Answer

(a) \( \mathrm{Zn(OH)_2 + H_2SO_4 \rightarrow ZnSO_4 + 2H_2O} \). (b) \(\mathrm{Zn(OH)_2}\) is the limiting reactant. (c) 0 mol \(\mathrm{Zn(OH)_2}\), 0.02885 mol \(\mathrm{H_2SO_4}\), and 0.0874 mol \(\mathrm{ZnSO_4}\).

Step 1: Write the Balanced Chemical Equation

The reaction between zinc hydroxide, \(\mathrm{Zn(OH)_2}\), and sulfuric acid, \(\mathrm{H_2SO_4}\), is a double displacement reaction. The products formed are zinc sulfate, \(\mathrm{ZnSO_4}\), and water as follows: \[ \mathrm{Zn(OH)_2 + H_2SO_4 \rightarrow ZnSO_4 + 2H_2O} \] This is the balanced chemical equation.

Step 2: Calculate Moles of Reactants

First, calculate the moles of \(\mathrm{Zn(OH)_2}\). Given mass is \(8.69\, \mathrm{g}\) and the molar mass of \(\mathrm{Zn(OH)_2}\) is approximately \(99.41\, \mathrm{g/mol}\). Using the formula \( n = \frac{m}{M} \), where \( m \) is mass, and \( M \) is molar mass: \[ n_{\mathrm{Zn(OH)_2}} = \frac{8.69}{99.41} \approx 0.0874 \text{ mol} \] Next, calculate the moles of \(\mathrm{H_2SO_4}\) using \( C = 0.750 \mathrm{M} \) and \( V = 0.155 \mathrm{L} \): \[ n_{\mathrm{H_2SO_4}} = C \times V = 0.750 \times 0.155 = 0.11625 \text{ mol} \]

Step 3: Identify the Limiting Reactant

To find the limiting reactant, compare the stoichiometric ratios. \(\mathrm{Zn(OH)_2 : H_2SO_4}\) is \(1:1\). Compare moles: \(0.0874\) mol \(\mathrm{Zn(OH)_2}\) with \(0.11625\) mol \(\mathrm{H_2SO_4}\). Since \(\mathrm{Zn(OH)_2}\) has fewer moles, it is the limiting reactant.

Step 4: Calculate Moles of Remaining and Produced Substances

Since \(\mathrm{Zn(OH)_2}\) is the limiting reactant, it will be completely consumed. The moles of \(\mathrm{Zn(OH)_2}\) are initially \(0.0874\) mol, so after the reaction it will be \(0\) mol. According to the reaction, \(0.0874\) mol of \(\mathrm{H_2SO_4}\) will react, leaving: \[ \text{Remaining } \mathrm{H_2SO_4} = 0.11625 - 0.0874 = 0.02885 \text{ mol} \] For \(\mathrm{ZnSO_4}\), the moles produced will equal the moles of \(\mathrm{Zn(OH)_2}\) that reacted: \[ n_{\mathrm{ZnSO_4}} = 0.0874 \text{ mol} \]

Key Concepts

Stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions. It involves calculations based on balanced chemical equations. This is crucial for determining how much reactant is needed or how much product can be formed.
The stoichiometric coefficients in a chemical equation reveal the ratios of the amount of each reactant and product.
For instance, the equation for the reaction between zinc hydroxide and sulfuric acid:

• \( \mathrm{Zn(OH)_2 + H_2SO_4 \rightarrow ZnSO_4 + 2H_2O} \)

The coefficients tell us that 1 mole of \( \mathrm{Zn(OH)_2} \) reacts with 1 mole of \( \mathrm{H_2SO_4} \) to produce 1 mole of \( \mathrm{ZnSO_4} \) and 2 moles of water. This helps in calculating quantities to use or expect in a reaction.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is entirely consumed first, halting the reaction as no more products can form.
To determine the limiting reactant, compare the mole ratio of reactants to the ratios in the balanced equation.
In our example, the reaction requires 1 mole of zinc hydroxide for every mole of sulfuric acid.
We have \(0.0874\) moles of \( \mathrm{Zn(OH)_2} \) and \(0.11625\) moles of \( \mathrm{H_2SO_4} \).
Because there are fewer moles of \( \mathrm{Zn(OH)_2} \), it runs out first and is the limiting reactant.
Once the limiting reactant is used up, no more \( \mathrm{ZnSO_4} \) or water can form.

Balanced Chemical Equation

A balanced chemical equation ensures the conservation of mass. It reflects the law of conservation of mass, stating that atoms cannot be created or destroyed in a chemical reaction.
In our scenario, the equation

• \( \mathrm{Zn(OH)_2 + H_2SO_4 \rightarrow ZnSO_4 + 2H_2O} \)

shows each element has the same number of atoms on both sides.
Balancing involves adjusting coefficients before compound formulas to get equal numbers of atoms for every element.

• 1 atom of Zn, 4 atoms of H, 2 atoms of O, and 1 atom of S on each side of the equation.

This is key to calculating how reactants convert to products, crucial for stoichiometry calculations.

Moles Calculation

Calculating moles helps in quantifying substances in reactions based on their mass and concentration. Moles are calculated differently for solids and solutions.
For solids, like zinc hydroxide here, calculate moles using molar mass:

• \( n = \frac{m}{M} = \frac{8.69}{99.41} \approx 0.0874 \text{ mol } \)

For solutions, moles are determined by multiplying molarity by volume in liters. Here, with sulfuric acid:

• \( n = C \times V = 0.750 \times 0.155 = 0.11625 \text{ mol } \)

Understanding moles helps determine amounts of reactants and products. Thus, it is essential for predicting reaction outcomes and calculating quantities needed or produced.