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Chapter 4: Problem 109

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

Expert verified
(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 by step solution

01

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.
02

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 \( V = 0.155 \mathrm{L} \): \[ n_{\mathrm{H_2SO_4}} = C \times V = 0.750 \times 0.155 = 0.11625 \text{ mol} \]
03

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.
04

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: \[ n_{\mathrm{ZnSO_4}} = 0.0874 \text{ mol} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Which of the following are redox reactions? For those that are, indicate which element is oxidized and which is reduced. For those that are not, indicate whether they are precipitation or neutralization reactions. (a) \(\mathrm{P}_{4}(s)+10 \mathrm{HClO}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$ 4 \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+10 \mathrm{HCl}(a q) $$ (b) \(\mathrm{Br}_{2}(l)+2 \mathrm{~K}(s) \longrightarrow 2 \mathrm{KBr}(s)\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g)\) (d) \(\mathrm{ZnCl}_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Zn}(\mathrm{OH})_{2}(s)+\) $$ 2 \mathrm{NaCl}(a q) $$

An \(8.65-g\) sample of an unknown group 2 metal hydroxide is dissolved in \(85.0 \mathrm{~mL}\) of water. An acid-base indicator is added and the resulting solution is titrated with \(2.50 \mathrm{M}\) \(\mathrm{HCl}(a q)\) solution. The indicator changes color, signaling that the equivalence point has been reached, after \(56.9 \mathrm{~mL}\) of the hydrochloric acid solution has been added. (a) What is the molar mass of the metal hydroxide? (b) What is the identity of the metal cation: \(\mathrm{Ca}^{2+}, \mathrm{Sr}^{2+}\), or \(\mathrm{Ba}^{2+}\) ?

Glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3},\) is a substance used extensively in the manufacture of cosmetic s, foodstuffs, antifreeze, and plastics. Glycerol is a water-soluble liquid with a density of \(1.2656 \mathrm{~g} / \mathrm{mL}\) at \(15^{\circ} \mathrm{C}\). Calculate the molarity of a solution of glycerol made by dissolving \(50.000 \mathrm{~mL}\) glycerol at \(15^{\circ} \mathrm{C}\) in enough water to make \(250.00 \mathrm{~mL}\) of solution.

Separate samples of a solution of an unknown salt are treated with dilute solutions of \(\mathrm{HBr}, \mathrm{H}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaOH}\). A precipitate forms in all three cases. Which of the following cations could be present in the unknown salt solution: \(\mathrm{K}^{+}, \mathrm{Pb}^{2+}, \mathrm{Ba}^{2+}\) ?

State whether each of the following statements is true or false. Justify your answer in each case. (a) \(\mathrm{NH}_{3}\) contains no OH \(^{-}\) ions, and yet its aqueous solutions are basic. (b) HF is a strong acid. (c) Although sulfuric acid is a strong electrolyte, an aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) contains more \(\mathrm{HSO}_{4}^{-}\) ions than \(\mathrm{SO}_{4}^{2-}\) ions.
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