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For each of the reactions, calculate how many grams of the

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 34P Chapter 8

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 34P

PROBLEM 34P

For each of the reactions, calculate how many grams of the product form when 17.8 g of the reactant in color completely reacts. Assume there is more than enough of the other reactant.

(a) Ca(s) + Cl2(g) → CaCl2(s)

(b) 2 K(s) + Br2(l) → 2 KBr(s)

(c) 4 Cr(s) + 3 O2(g) → 2 Cr2O3(s)

(d) 2 Sr(s) + O2(g) → 2 SrO(s)

Step-by-Step Solution:

Solution 34P

Here, we are going to calculate the mass of product when 17.8 g of the reactant in each case completely reacts.

Step 1:

Ca(s) + Cl2(g) → CaCl2(s)

Here, the coloured reactant is Chlorine.

Given, mass of Cl2 reacted = 17.8 g

Molar mass of Cl2 = 2 x 35.453 = 70.906 g/mol

Therefore, number of moles of Cl2 = given mass / molar mass

                                   =

                                   = 0.25 mol.

From the equation given above, it is seen that 1 mol of Cl2 produces 1 mol of CaCl2.

Therefore, 0.25 mol of Cl2 will produce 0.25 mol of CaCl2.

Now, 1 mol CaCl2 = molar mass of CaCl2

Therefore, 0.25 mol of CaCl2 = 0.25 x molar mass of CaCl2

                                = 0.25 x 110.98 g/mol

                                = 27.7 g

Thus 27.7 g CaCl2 will be produced from reaction (a).

Step 2:

b) 2K(s) + Br2(l) → 2KBr(s)

Here, the coloured reactant is Bromine.

Given, mass of Br2 reacted = 17.8 g

Molar mass of Br2 = 2 x 79.904 = 159.8 g/mol

Therefore, number of moles of Br2 = given mass / molar mass

                                   =

                                   = 0.11 mol.

From the equation given above, it is seen that 1 mol of Br2 produces 2 mol of KBr.

Therefore, 0.11 mol of Br2 will produce (0.11 x 2 = 0.22) mol of KBr.

Now, 1 mol KBr = molar mass of KBr

Therefore, 0.22 mol of KBr = 0.22 x molar mass of KBr

                                = 0.22 x 119.0 g/mol

                                = 26.18 g

Thus 26.18 g KBr will be produced from reaction (b).

Step 3 of 4

Chapter 8, Problem 34P is Solved
Step 4 of 4

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. Since the solution to 34P from 8 chapter was answered, more than 1032 students have viewed the full step-by-step answer. This full solution covers the following key subjects: reactant, grams, calculate, color, completely. This expansive textbook survival guide covers 19 chapters, and 2045 solutions. The answer to “For each of the reactions, calculate how many grams of the product form when 17.8 g of the reactant in color completely reacts. Assume there is more than enough of the other reactant.(a) Ca(s) + Cl2(g) ? CaCl2(s)(b) 2 K(s) + Br2(l) ? 2 KBr(s)(c) 4 Cr(s) + 3 O2(g) ? 2 Cr2O3(s)(d) 2 Sr(s) + O2(g) ? 2 SrO(s)” is broken down into a number of easy to follow steps, and 60 words. The full step-by-step solution to problem: 34P from chapter: 8 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5.

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