A sample of 8.69 g of Zn(OH) 2 is added to 155.0 mL of 0.750 M H 2 SO 4 . (b) Which is the limiting reactant in the reaction?

Welcome back everyone. The mixture of sodium bicarbonate and acetic acid produces bubbles. According to the following equation where sodium bicarbonate solid reacts with acetic acid to produce one mole of carbon dioxide gas, one mole of liquid water and one mole of sodium acetate determined the limiting reactant. If 1.50 g of sodium bicarbonate and 1.50 g of acetic acid were allowed to react. So we're going to begin by finding our moles of our product. Carbon dioxide from each of our reactant in this case will begin with sodium bicarbonate. So we know that we have From the prompt g of sodium bicarbonate that are reacting. And we want to go from grams of sodium bicarbonate, two moles of sodium bicarbonate in the numerator. And so here we can utilize our molar mass of sodium bicarbonate from the periodic table, which will see for one mole of sodium bicarbonate, Has an equivalent of 84.007 g of sodium bicarbonate. So canceling out grams of sodium bicarbonate, we are at moles and we need to get from molds of our reactant two moles of our product carbon dioxide in the numerator. So in our denominator will plug in the moles of sodium bicarbonate and we want to refer to our molar ratio from our equation which is actually balanced as written. And we can see that we have for one mole of our sodium bicarbonate reactant, one mole of carbon dioxide gas produced. So that's a 1-1 molar ratio, canceling out our moles of sodium bicarbonate. Were left with moles of our product, which is what we wanted to determine. And this gives us a value of 0.1786 moles of carbon dioxide produced. Now we're comparing this to our moles of carbon dioxide produced from our second reactant being acetic acid, ch three C 00 H. So beginning with our mass of acetic acid. And sorry, this is 1.50 here. So that is also the same as our sodium bicarbonate. 1.50 g of acetic acid. We want to multiply by our molar mass as a conversion factor where for one mole of acetic acid, Our periodic table gives us a mass of 60.052 g of acetic acid. So canceling out grams of acetic acid. We're going to go from molds of our second reactant, molds of acetic acid, two moles of our product carbon dioxide And referring to our balanced equation, we have one mole of acetic acid which produces one mole of carbon dioxide. So also a 1 to 1 molar ratio canceling out moles of acetic acid. We're left with most of our product C. 02 and we're going to get a result of 0.2498 moles of carbon dioxide produced from acetic acid. And because we see that are moles of carbon dioxide produced from bicarbonate is lower than our moles of carbon dioxide made from acetic acid. We would say that therefore, sodium bicarbonate is our limiting reactant. And so this statement would be our final answer to complete this example, corresponding to choice, be in the multiple choice as the correct answer. So I hope this made sense and let us know if you have any questions.

Ch.4 - Reactions in Aqueous Solution

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Brown 14th Edition

Ch.4 - Reactions in Aqueous Solution

Problem 109b

Chapter 4, Problem 109b

A sample of 8.69 g of Zn(OH)₂ is added to 155.0 mL of 0.750 M H₂SO₄. (b) Which is the limiting reactant in the reaction?

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First, write the balanced chemical equation for the reaction between zinc hydroxide (Zn(OH)₂) and sulfuric acid (H₂SO₄). The balanced equation is: Zn(OH)₂ + H₂SO₄ → ZnSO₄ + 2H₂O.

Calculate the number of moles of Zn(OH)₂ using its molar mass. The molar mass of Zn(OH)₂ is approximately 99.41 g/mol. Use the formula: moles = mass / molar mass.

Calculate the number of moles of H₂SO₄ using its concentration and volume. The concentration is given as 0.750 M and the volume is 155.0 mL. Convert the volume to liters and use the formula: moles = concentration × volume.

Determine the stoichiometric ratio from the balanced equation. According to the equation, 1 mole of Zn(OH)₂ reacts with 1 mole of H₂SO₄. Compare the moles of each reactant to find the limiting reactant.

Identify the limiting reactant by comparing the calculated moles of Zn(OH)₂ and H₂SO₄. The reactant with fewer moles than required by the stoichiometric ratio is the limiting reactant.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced chemical equation. It involves using mole ratios derived from the coefficients of the balanced equation to determine how much of each reactant is needed or how much product can be formed. Understanding stoichiometry is essential for identifying limiting reactants, as it allows for the comparison of the amounts of reactants available.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed. Once the limiting reactant is used up, the reaction cannot proceed further, even if other reactants are still available. Identifying the limiting reactant is crucial for calculating yields and understanding the efficiency of a reaction.

Molarity and Volume Calculations

Molarity (M) is a measure of concentration defined as the number of moles of solute per liter of solution. To determine the amount of a reactant in a solution, one can use the formula: moles = molarity × volume (in liters). In this question, calculating the moles of H2SO4 using its molarity and the given volume is necessary to compare it with the moles of Zn(OH)2 to find the limiting reactant.

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Molar Mass Calculation Example

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(b) An elemental analysis of the acid indicates that it is composed of 5.89% H, 70.6% C, and 23.5% O by mass. What is its molecular formula?

Textbook Question

The discovery of hafnium, element number 72, provideda controversial episode in chemistry. G. Urbain, a Frenchchemist, claimed in 1911 to have isolated an elementnumber 72 from a sample of rare earth (elements 58–71)compounds. However, Niels Bohr believed that hafniumwas more likely to be found along with zirconium thanwith the rare earths. D. Coster and G. von Hevesy, workingin Bohr's laboratory in Copenhagen, showed in 1922 thatelement 72 was present in a sample of Norwegian zircon,an ore of zirconium. (The name hafnium comes from theLatin name for Copenhagen, Hafnia). (c) Solid zirconiumdioxide, ZrO2, reacts with chlorine gas in the presenceof carbon. The products of the reaction are ZrCl4 and twogases, CO2 and CO in the ratio 1:2. Write a balanced chemicalequation for the reaction.

Textbook Question

A sample of 8.69 g of Zn(OH)₂ is added to 155.0 mL of 0.750 M H₂SO₄. (c) How many moles of Zn(OH)₂, H₂SO₄, ZnSO₄ are present after the reaction is complete?

Textbook Question

In 2014, a major chemical leak at a facility in West Virginia released 28,390 L of MCHM (4-methylcyclohexylmethanol, C8H16O) into the Elk River. The density of MCHM is 0.9074 g/mL. (a) Calculate the initial molarity of MCHM in the river, assuming that the first part of the river is 2.00 m deep, 90.0 m wide, and 90.0 m long.

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Ritalin is the trade name of a drug, methylphenidate, used to treat attention-deficit/hyperactivity disorder in young adults. The chemical structure of methylphenidate is (c) Ritalin has a half-life of 3 hours in the blood, which means that after 3 hours the concentration in the blood has decreased by half of its initial value. For the man in part (b), what is the concentration of Ritalin in his blood after 6 hours?

A sample of 8.69 g of Zn(OH)2 is added to 155.0 mL of 0.750 M H2SO4. (b) Which is the limiting reactant in the reaction?

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Key Concepts

Stoichiometry

Limiting Reactant

Molarity and Volume Calculations

A sample of 8.69 g of Zn(OH)₂ is added to 155.0 mL of 0.750 M H₂SO₄. (b) Which is the limiting reactant in the reaction?