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 + H_2SO_4 → ZnSO_4 + 2H_2O.

Calculate the number of moles of Zn(OH)2. Use the molar mass of Zn(OH)2, which is approximately 99.41 g/mol, to convert grams to moles: moles of Zn(OH)2 = 8.69 g / 99.41 g/mol.

Calculate the number of moles of H2SO4. Use the molarity and volume of the solution to find moles: moles of H2SO4 = 0.750 M * 0.155 L.

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 calculated moles of Zn(OH)2 and H2SO4 to determine the limiting reactant. The reactant with fewer moles based on the stoichiometric ratio is the limiting reactant.

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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?

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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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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?