Considering the chemical equation below: C2H4B12 + 3 02→ 2 CO2 + 2 H20 + Br2 What is the maximum amount of bromine (Br2) that can be obtained from mixing 76.4 g of C2H4Br2 with 49.1 g of 02 (Mw of C2H4B12 = 187.86 %3D amu, O2 = 32.00 amu and Br2 = 159.81 amu)?

Considering the chemical equation below: C2H4B12 + 3 02→ 2 CO2 + 2 H20 + Br2 What is the maximum amount of bromine (Br2) that can be obtained from mixing 76.4 g of C2H4Br2 with 49.1 g of 02 (Mw of C2H4B12 = 187.86 %3D amu, O2 = 32.00 amu and Br2 = 159.81 amu)?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Considering the chemical equation
below:
C2НдВr2 + 3 02 2 СО2 + 2 H2О +
Br2
What is the maximum amount of
bromine (Br2) that can be obtained
from mixing 76.4 g of C2HĄBr2 with
49.1 g of 02 (Mw of C2H4B12 = 187.86
%3D
amu, O2 = 32.00 amu and Br2 =
159.81 amu)?
%D
68.7 g
O 21.7 g
81.7 g
16.2 g
O 65.0 g

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