Solution: Calculating the Amount of Product Formed from a

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Problem 2PE Chapter 3.19SE

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 2PE

Calculating the Amount of Product Formed from a Limiting Reactant The reaction is used to produce electricity in a hydrogen fuel cell. Suppose a fuel cell contains 150 g of H2(g) and 1500 g of O2(g) (each measured to two significant figures). How many grams of water can form? When a 2.00-g strip of zinc metal is placed in an aqueous solution containing 2.50 g of silver nitrate, the reaction is (a) Which reactant is limiting? (b) How many grams of Ag form? (c) How many grams of Zn(NO3)2 form? (d) How many grams of the excess reactant are left at the end of the reaction?

Step-by-Step Solution:

Answer: We have to calculate the amount of product formed from a limiting reactant The reaction is 2H2(g)+O2(g)2H2O(g). Step 1 We are asked to calculate the amount of a product, given the amounts of two reactants, so this is a limiting reactant problem. Step 2 To identify the limiting reactant, we can calculate the number of moles of each reactant and compare their ratio with the ratio of coefficients in the balanced equation. We then use the quantity of the limiting reactant to calculate the mass of water that forms. Step 3 From the balanced equation, we have the stoichiometric relations . Here 2 moles of hydrogen reacts with one mole of hydrogen to give 2 moles of H2O . Using the molar mass of each substance, we calculate the number of moles of each reactant: Step 4 Moles H2 = (150 g H2)*(1 mol H2)/( 2.02 g H2 ) = 74 mol H2. Moles O2=(1500 gO2)*(1 mol O2)/(16.0 g O2)=47 mol O2. The coefficients in the balanced equation indicate that the reaction requires 2 mol of H2 for every 1 mol of O2. Therefore, for all the O2 to completely react, we would need 2 * 47 = 94 mol of H2. Since there are only 74 mol of H2, all of the O2 cannot react, so it is the excess reactant, and H2 must be the limiting reactant. (Notice that the limiting reactant is not necessarily the one present in the lowest amount.) We use the given quantity of H2 (the limiting reactant) to calculate the quantity of water formed. We could begin this calculation with the given H2 mass, 150 g, but we can save a step by starting with the moles of H2, 74 mol, we just calculated: Grams H2O=1.3 *10^2 g of H2O. Step 5 The magnitude of the answer seems reasonable based on the amounts of the reactants. The units are correct, and the number of significant figures (two) corresponds to those in the values given in the problem statement. Comment The quantity of the limiting reactant, H2, can also be used to determine the quantity of O2 used: Grams O2 =(74 mol H2)*( 1 mol O2/ 2 mol H2 )*( 32.0 g O2 )/(1 mol O2) =( 1.2 * 10^3g O2). The mass of O2 remaining at the end of the reaction equals the starting amount minus the amount consumed: 1500 g - 1200 g = 300 g.

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Chapter 3.19SE, Problem 2PE is Solved
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Textbook: Chemistry: A Molecular Approach
Edition: 3rd
Author: Nivaldo J. Tro
ISBN: 9780321809247

This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3rd. The answer to “Calculating the Amount of Product Formed from a Limiting Reactant The reaction is used to produce electricity in a hydrogen fuel cell. Suppose a fuel cell contains 150 g of H2(g) and 1500 g of O2(g) (each measured to two significant figures). How many grams of water can form? When a 2.00-g strip of zinc metal is placed in an aqueous solution containing 2.50 g of silver nitrate, the reaction is (a) Which reactant is limiting? (b) How many grams of Ag form? (c) How many grams of Zn(NO3)2 form? (d) How many grams of the excess reactant are left at the end of the reaction?” is broken down into a number of easy to follow steps, and 106 words. The full step-by-step solution to problem: 2PE from chapter: 3.19SE was answered by Sieva Kozinsky, our top Chemistry solution expert on 02/22/17, 04:35PM. Since the solution to 2PE from 3.19SE chapter was answered, more than 312 students have viewed the full step-by-step answer. This full solution covers the following key subjects: grams, reactant, reaction, form, Cell. This expansive textbook survival guide covers 82 chapters, and 9464 solutions. Chemistry: A Molecular Approach was written by Sieva Kozinsky and is associated to the ISBN: 9780321809247.

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